[proofplan] The Schwartz kernel of an oscillatory operator is the oscillatory integral $K(x,y)=(2\pi)^{-n}\int e^{i(x-y)\cdot\xi}A(x,y,\xi)\,d\mathcal{L}^n(\xi)$; we first record that lowering the order of the amplitude raises the smoothness of this kernel, so that an amplitude of order $-\infty$ produces a smoothing operator. After cutting $A$ off to a neighbourhood of the diagonal (the off-diagonal part is automatically smoothing), we Taylor expand $A(x,y,\xi)$ in $y$ about $y=x$ to order $N$ with integral remainder. Using $\partial_\xi^\alpha e^{i(x-y)\cdot\xi}=i^{|\alpha|}(x-y)^\alpha e^{i(x-y)\cdot\xi}$, each monomial $(y-x)^\alpha$ is converted into $\xi$-derivatives of the phase, and [integration by parts](/theorems/2098) in $\xi$ transfers these derivatives onto the amplitude, producing exactly the operators $\operatorname{Op}_L(b_j)$ for $|\alpha|=j$, plus a remainder operator whose amplitude has order $m-N$. We verify $b_j\in S^{m-j}$ directly from the amplitude estimates, sum the $b_j$ into a single symbol $a\in S^m$ with $a-\sum_{j<N}b_j\in S^{m-N}$ by a Borel-type construction, and conclude that $I_A-\operatorname{Op}_L(a)$ has a kernel of class $C^k$ for every $k$, hence is smoothing. [/proofplan] [step:Represent the operators by Schwartz kernels and bound kernel regularity by the order of the amplitude] Throughout, oscillatory integrals are understood in the standard regularized sense: fixing $\rho\in C_c^\infty(\mathbb{R}^n)$ with $\rho(0)=1$, every integral $\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}B(x,y,\xi)\,d\mathcal{L}^n(\xi)$ below means $\lim_{\varepsilon\to0^+}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}\rho(\varepsilon\xi)B(x,y,\xi)\,d\mathcal{L}^n(\xi)$, the limit existing in $\mathcal{D}'(U\times U)$ and being independent of $\rho$. For each $\varepsilon>0$ the integrand is absolutely integrable, so the pointwise algebraic identities used below (Taylor expansion in $y$, the phase identity, and integration by parts in $\xi$) hold for the $\varepsilon$-regularized integrals and pass to the limit; we apply them without further comment. For an amplitude $B$ of order $\mu$, the operator $I_B$ has Schwartz kernel given by the map $K_B:U\times U\to\mathbb{C}$, \begin{align*} K_B(x,y)=(2\pi)^{-n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}B(x,y,\xi)\,d\mathcal{L}^n(\xi), \end{align*} so that $I_Bu(x)=\int_U K_B(x,y)\,u(y)\,d\mathcal{L}^n(y)$ for $u\in C_c^\infty(U)$. [claim:An amplitude of order $\mu<-n-k$ has a kernel of class $C^k$] If $B$ is an amplitude of order $\mu$ and $k\in\mathbb{N}_0$ satisfies $\mu<-n-k$, then $K_B\in C^k(U\times U)$. Consequently, if $B$ is an amplitude of order $-\infty$, then $K_B\in C^\infty(U\times U)$ and $I_B$ is smoothing. [/claim] [proof] Fix compact sets $K,L\subset U$ and multi-indices $p,q\in\mathbb{N}_0^n$ with $|p|+|q|\le k$. Since $\partial_{x_i}e^{i(x-y)\cdot\xi}=i\xi_i\,e^{i(x-y)\cdot\xi}$ and $\partial_{y_i}e^{i(x-y)\cdot\xi}=-i\xi_i\,e^{i(x-y)\cdot\xi}$, the Leibniz rule gives, for the (absolutely convergent) $\varepsilon$-regularized integrals, \begin{align*} \partial_x^p\partial_y^q\!\left[e^{i(x-y)\cdot\xi}B(x,y,\xi)\right]=\sum_{\substack{p'+p''=p, q'+q''=q}}\binom{p}{p'}\binom{q}{q'}(i\xi)^{p'}(-i\xi)^{q'}\,e^{i(x-y)\cdot\xi}\,\partial_x^{p''}\partial_y^{q''}B(x,y,\xi). \end{align*} Each summand is bounded on $K\times L\times\mathbb{R}^n$ by $C\,|\xi|^{|p'|+|q'|}\langle\xi\rangle^{\mu}\le C\,\langle\xi\rangle^{\mu+|p|+|q|}\le C\,\langle\xi\rangle^{\mu+k}$, using the amplitude estimate for $\partial_x^{p''}\partial_y^{q''}B$ (order $\mu$, no $\xi$-derivatives) and $|\xi|^{|p'|+|q'|}\le\langle\xi\rangle^{|p'|+|q'|}$. Because $\mu+k<-n$, the function $\xi\mapsto\langle\xi\rangle^{\mu+k}$ is integrable on $\mathbb{R}^n$, so the bound is an $\varepsilon$-independent integrable majorant. By the [Dominated Convergence Theorem](/theorems/4) the regularized integrals converge as $\varepsilon\to0^+$ to the absolutely convergent integral, and by the [Leibniz Integral Rule](/theorems/831) (differentiation under the integral sign, whose hypotheses are exactly the existence of this majorant) the function $K_B$ is $|p|+|q|$ times continuously differentiable with \begin{align*} \partial_x^p\partial_y^q K_B(x,y)=(2\pi)^{-n}\int_{\mathbb{R}^n}\partial_x^p\partial_y^q\!\left[e^{i(x-y)\cdot\xi}B(x,y,\xi)\right]d\mathcal{L}^n(\xi). \end{align*} As $K,L$ were arbitrary compacts and $|p|+|q|\le k$ arbitrary, $K_B\in C^k(U\times U)$. If $B$ is of order $-\infty$, it is of order $\mu$ for every $\mu$; given any $k$, choosing $\mu<-n-k$ shows $K_B\in C^k(U\times U)$. Hence $K_B\in C^\infty(U\times U)$, and $I_B$ is smoothing by definition. [/proof] [/step] [step:Localize the amplitude to a neighbourhood of the diagonal where segments stay in $U$] The segment $[x,y]=\{x+t(y-x):t\in[0,1]\}$ need not lie in $U$ for arbitrary $x,y\in U$, yet the Taylor remainder in Step 3 requires it. We remove this obstruction. Choose $\chi\in C^\infty(U\times U)$ with $0\le\chi\le1$, equal to $1$ on a neighbourhood $\mathcal{N}$ of the diagonal $\Delta=\{(x,x):x\in U\}$, and supported in a set $\mathcal{V}$ such that $[x,y]\subset U$ for every $(x,y)\in\mathcal{V}$ and $\overline{\mathcal V}$ meets each $K\times K$ ($K\subset U$ compact) in a compact subset of $U\times U$. Such $\chi$ exists: cover $\Delta$ by balls $B(x_0,r)\times B(x_0,r)$ with $\overline{B}(x_0,2r)\subset U$ (these are convex, so segments stay inside), take a locally finite subcover, and let $\chi$ be a smooth function that is $1$ near $\Delta$ and supported in their union. Write $A=\chi A+(1-\chi)A$, where $\chi A$ and $(1-\chi)A$ are amplitudes of order $m$ (the cutoff is $\xi$-independent, so the order is unchanged). On $\operatorname{supp}(1-\chi)$ we have $(x,y)\notin\mathcal N$, hence $|x-y|\ge\delta_{K}>0$ on each $K\times K$. [claim:The off-diagonal operator $I_{(1-\chi)A}$ is smoothing] The kernel $K_{(1-\chi)A}\in C^\infty(U\times U)$. [/claim] [proof] Fix compacts $K,L\subset U$; on $K\times L$, points of $\operatorname{supp}(1-\chi)$ satisfy $|x-y|\ge\delta>0$. Since $(-\Delta_\xi)e^{i(x-y)\cdot\xi}=|x-y|^2\,e^{i(x-y)\cdot\xi}$, for $|x-y|>0$ and any $M\in\mathbb{N}$ we have $e^{i(x-y)\cdot\xi}=|x-y|^{-2M}(-\Delta_\xi)^M e^{i(x-y)\cdot\xi}$. Integrating by parts $M$ times in $\xi$ (justified on the $\varepsilon$-regularized integrals, with no boundary contribution), \begin{align*} K_{(1-\chi)A}(x,y)=(2\pi)^{-n}\,|x-y|^{-2M}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}(-\Delta_\xi)^M\!\big[(1-\chi(x,y))A(x,y,\xi)\big]\,d\mathcal{L}^n(\xi). \end{align*} The amplitude $(-\Delta_\xi)^M[(1-\chi)A]$ has order $m-2M$. Given any $k$, choose $M$ with $m-2M<-n-k$; the previous claim's argument (applied on the [open set](/page/Open%20Set) $\{|x-y|>\delta/2\}$, where $|x-y|^{-2M}$ is smooth) shows the integral defines a $C^k$ function there, and multiplying by the smooth factor $|x-y|^{-2M}$ keeps it $C^k$. As $k$ is arbitrary and $\operatorname{supp}(1-\chi)\subset\{|x-y|\ge\delta\}$, we get $K_{(1-\chi)A}\in C^\infty(U\times U)$. [/proof] Thus $I_A-I_{\chi A}=I_{(1-\chi)A}$ is smoothing, and it suffices to construct the symbol $a$ for the localized amplitude $A_0:=\chi A$. Finally, since $\chi\equiv1$ on the neighbourhood $\mathcal N$ of $\Delta$, all $y$-derivatives of $\chi$ vanish on $\Delta$, so by the Leibniz rule $\partial_y^\alpha A_0(x,y,\xi)\big|_{y=x}=\partial_y^\alpha A(x,y,\xi)\big|_{y=x}$ for every $\alpha$; the coefficients $b_j$ computed from $A_0$ coincide with those computed from $A$. [/step] [step:Taylor expand the localized amplitude in $y$ about the diagonal with integral remainder] Fix $N\in\mathbb{N}$. For fixed $(x,\xi)$ with $(x,y)\in\operatorname{supp}\chi$, the map $y\mapsto A_0(x,y,\xi)$ is smooth on the convex set containing the segment $[x,y]\subset U$. By [Taylor's Theorem with Integral Remainder](/theorems/189), expanding in $y$ about the base point $x$ to order $N$, \begin{align*} A_0(x,y,\xi)=\sum_{|\alpha|<N}\frac{(y-x)^\alpha}{\alpha!}\,\partial_y^\alpha A_0(x,x,\xi)+R_N(x,y,\xi), \end{align*} with remainder \begin{align*} R_N(x,y,\xi)=\sum_{|\alpha|=N}\frac{N}{\alpha!}(y-x)^\alpha\int_0^1(1-t)^{N-1}\,\partial_y^\alpha A_0\big(x,\,x+t(y-x),\,\xi\big)\,d\mathcal{L}^1(t). \end{align*} For $|\alpha|=N$ define the map $\widetilde{A}_\alpha:U\times U\times\mathbb{R}^n\to\mathbb{C}$, \begin{align*} \widetilde{A}_\alpha(x,y,\xi)=\frac{N}{\alpha!}\int_0^1(1-t)^{N-1}\,\partial_y^\alpha A_0\big(x,\,x+t(y-x),\,\xi\big)\,d\mathcal{L}^1(t), \end{align*} so that $R_N=\sum_{|\alpha|=N}(y-x)^\alpha\,\widetilde{A}_\alpha$, and set $g_\alpha(x,\xi):=\partial_y^\alpha A_0(x,y,\xi)\big|_{y=x}$ for the diagonal coefficients. Each $\widetilde A_\alpha$ is an amplitude of order $m$. Indeed, for compacts $K,L\subset U$ with $K\times L\subset\operatorname{supp}\chi$, the points $x+t(y-x)$ ($t\in[0,1]$, $(x,y)\in K\times L$) lie in a fixed compact $M\subset U$; differentiating under the integral sign by the [Leibniz Integral Rule](/theorems/831) and applying the chain rule in the middle slot, every derivative $\partial_x^\beta\partial_y^\gamma\partial_\xi^\delta\widetilde A_\alpha$ is a finite combination of integrals of $(\partial_x^{\beta'}\partial_y^{\gamma'}\partial_\xi^\delta\partial_y^\alpha A_0)(x,x+t(y-x),\xi)$ against the bounded weight $t^{|\gamma''|}(1-t)^{N-1}$, each integrand bounded by $C\langle\xi\rangle^{m-|\delta|}$ on $K\times L\times\mathbb{R}^n$ via the order-$m$ estimate for $A_0$. Integrating in $t\in[0,1]$ preserves the bound, so $|\partial_x^\beta\partial_y^\gamma\partial_\xi^\delta\widetilde A_\alpha|\le C\langle\xi\rangle^{m-|\delta|}$. [/step] [step:Convert the monomials $(y-x)^\alpha$ into $\xi$-derivatives and integrate by parts] We feed the expansion of Step 3 into the kernel of $I_{A_0}$. The basic identity is \begin{align*} \partial_\xi^\alpha e^{i(x-y)\cdot\xi}=\big(i(x-y)\big)^\alpha e^{i(x-y)\cdot\xi}=i^{|\alpha|}(x-y)^\alpha e^{i(x-y)\cdot\xi}, \end{align*} hence, since $(y-x)^\alpha=(-1)^{|\alpha|}(x-y)^\alpha$ and $(-1)^{|\alpha|}i^{-|\alpha|}=i^{|\alpha|}$, \begin{align*} (y-x)^\alpha e^{i(x-y)\cdot\xi}=i^{|\alpha|}\,\partial_\xi^\alpha e^{i(x-y)\cdot\xi}. \end{align*} For each $\alpha$ with $|\alpha|<N$ the corresponding term of $I_{A_0}$ has kernel contribution \begin{align*} (2\pi)^{-n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}\frac{(y-x)^\alpha}{\alpha!}g_\alpha(x,\xi)\,d\mathcal{L}^n(\xi) =(2\pi)^{-n}\frac{i^{|\alpha|}}{\alpha!}\int_{\mathbb{R}^n}\big[\partial_\xi^\alpha e^{i(x-y)\cdot\xi}\big]g_\alpha(x,\xi)\,d\mathcal{L}^n(\xi). \end{align*} Integrating by parts $|\alpha|$ times in $\xi$ (no boundary terms on the $\varepsilon$-regularized integrals) moves $\partial_\xi^\alpha$ onto $g_\alpha$ with sign $(-1)^{|\alpha|}$: \begin{align*} =(2\pi)^{-n}\frac{(-1)^{|\alpha|}i^{|\alpha|}}{\alpha!}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}\,\partial_\xi^\alpha g_\alpha(x,\xi)\,d\mathcal{L}^n(\xi). \end{align*} Now $(-1)^{|\alpha|}i^{|\alpha|}=(-i)^{|\alpha|}$, and $(-i)^{|\alpha|}\partial_\xi^\alpha g_\alpha=\partial_\xi^\alpha\big[(-i)^{|\alpha|}\partial_y^\alpha A_0\big]\big|_{y=x}=\partial_\xi^\alpha D_y^\alpha A_0(x,y,\xi)\big|_{y=x}$, since the constant $(-i)^{|\alpha|}$ and the operators $\partial_\xi^\alpha$, evaluation at $y=x$ all commute. Therefore the $\alpha$-term equals the kernel of $\operatorname{Op}_L(c_\alpha)$, where the map $c_\alpha:U\times\mathbb{R}^n\to\mathbb{C}$ is \begin{align*} c_\alpha(x,\xi)=\frac{1}{\alpha!}\,\partial_\xi^\alpha D_y^\alpha A_0(x,y,\xi)\big|_{y=x}. \end{align*} Summing over $|\alpha|=j$ gives exactly $\sum_{|\alpha|=j}c_\alpha=b_j$ (computed from $A_0$, which equals $b_j$ computed from $A$ by Step 2). The remainder term, treated identically with $|\alpha|=N$ and the amplitude $\widetilde A_\alpha$ in place of $g_\alpha$, yields the kernel of an operator $I_{B_N}$ whose amplitude is \begin{align*} B_N(x,y,\xi)=\sum_{|\alpha|=N}(-i)^{|\alpha|}\,\partial_\xi^\alpha\widetilde A_\alpha(x,y,\xi). \end{align*} Since each $\widetilde A_\alpha$ is an amplitude of order $m$ (Step 3) and $|\alpha|=N$, the amplitude $B_N$ has order $m-N$. We have shown the exact decomposition \begin{align*} I_{A_0}=\operatorname{Op}_L\Big(\sum_{j=0}^{N-1}b_j\Big)+I_{B_N},\qquad B_N\ \text{an amplitude of order } m-N. \end{align*} [guided] The whole step is engineered to turn the variable $y$ — which the left quantization $\operatorname{Op}_L$ is not allowed to contain — into the frequency variable $\xi$, which it is. The obstruction is the factor $(y-x)^\alpha$ produced by [Taylor's theorem](/theorems/827). Why can we hope to absorb it? Because the phase $e^{i(x-y)\cdot\xi}$ trades position differences for frequency derivatives: differentiating the phase in $\xi$ brings down exactly $i(x-y)$. Let us make this precise. Differentiating $e^{i(x-y)\cdot\xi}$ in $\xi$ componentwise, $\partial_{\xi_r}e^{i(x-y)\cdot\xi}=i(x_r-y_r)e^{i(x-y)\cdot\xi}$, and iterating, \begin{align*} \partial_\xi^\alpha e^{i(x-y)\cdot\xi}=i^{|\alpha|}(x-y)^\alpha e^{i(x-y)\cdot\xi}. \end{align*} We want $(y-x)^\alpha$, not $(x-y)^\alpha$; the sign flip is $(y-x)^\alpha=(-1)^{|\alpha|}(x-y)^\alpha$. Combining, and using $i^{-|\alpha|}(-1)^{|\alpha|}=i^{|\alpha|}$, \begin{align*} (y-x)^\alpha e^{i(x-y)\cdot\xi}=i^{|\alpha|}\,\partial_\xi^\alpha e^{i(x-y)\cdot\xi}. \end{align*} Now take the $\alpha$-term ($|\alpha|<N$) of the kernel of $I_{A_0}$. The Taylor coefficient $g_\alpha(x,\xi)=\partial_y^\alpha A_0(x,y,\xi)|_{y=x}$ depends only on $(x,\xi)$ — that is the reason we expanded around the diagonal $y=x$ — so it factors out of the $y$-behaviour: \begin{align*} (2\pi)^{-n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}\frac{(y-x)^\alpha}{\alpha!}g_\alpha(x,\xi)\,d\mathcal{L}^n(\xi) =(2\pi)^{-n}\frac{i^{|\alpha|}}{\alpha!}\int_{\mathbb{R}^n}\big[\partial_\xi^\alpha e^{i(x-y)\cdot\xi}\big]g_\alpha(x,\xi)\,d\mathcal{L}^n(\xi). \end{align*} The phase now carries the derivative $\partial_\xi^\alpha$; we move it onto the symbol by [integration by parts](/theorems/210) in $\xi$. Each integration introduces a sign $-1$, and there are no boundary terms because we work with the $\varepsilon$-regularized integrals, whose integrands have compact $\xi$-support; after $|\alpha|$ steps, \begin{align*} =(2\pi)^{-n}\frac{(-1)^{|\alpha|}i^{|\alpha|}}{\alpha!}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}\,\partial_\xi^\alpha g_\alpha(x,\xi)\,d\mathcal{L}^n(\xi). \end{align*} Here is where the convention $D_y^\alpha=(-i)^{|\alpha|}\partial_y^\alpha$ earns its place: the prefactor is $(-1)^{|\alpha|}i^{|\alpha|}=(-i)^{|\alpha|}$, and pulling this constant into the $y$-derivative converts $\partial_y^\alpha$ into $D_y^\alpha$, \begin{align*} (-i)^{|\alpha|}\partial_\xi^\alpha g_\alpha(x,\xi)=\partial_\xi^\alpha D_y^\alpha A_0(x,y,\xi)\big|_{y=x}. \end{align*} This is precisely $\alpha!\,c_\alpha(x,\xi)$, so the $\alpha$-term is the kernel of $\operatorname{Op}_L(c_\alpha)$ — a genuine left quantization, with $y$ eliminated. Summing over $|\alpha|=j$ collects $c_\alpha$ into $b_j$, reproducing the formula in the statement and confirming the sign convention is consistent with the phase $e^{i(x-y)\cdot\xi}$. What about the remainder $R_N=\sum_{|\alpha|=N}(y-x)^\alpha\widetilde A_\alpha$? The factor $(y-x)^\alpha$ is handled by the very same identity and integration by parts; the only difference is that $\widetilde A_\alpha$ still depends on $y$, so we obtain not a left symbol but a new amplitude $B_N=\sum_{|\alpha|=N}(-i)^{|\alpha|}\partial_\xi^\alpha\widetilde A_\alpha$. The crucial gain is in the order: applying $\partial_\xi^\alpha$ with $|\alpha|=N$ to an amplitude of order $m$ lowers its order to $m-N$. Thus the cost of stopping the Taylor expansion at order $N$ is a remainder operator $I_{B_N}$ whose amplitude is $N$ orders better. This is the mechanism that will make the difference smoothing once $N\to\infty$. [/guided] [/step] [step:Verify that the expansion coefficients satisfy $b_j\in S^{m-j}$] It suffices to show each $c_\alpha\in S^{m-j}(U\times\mathbb{R}^n)$ for $|\alpha|=j$, since $b_j=\sum_{|\alpha|=j}c_\alpha$ is a finite sum. Write \begin{align*} c_\alpha(x,\xi)=\frac{(-i)^{|\alpha|}}{\alpha!}\,\big[\partial_\xi^\alpha\partial_y^\alpha A_0\big](x,x,\xi), \end{align*} the right-hand side meaning $\partial_\xi^\alpha\partial_y^\alpha A_0$ evaluated with both the first and second arguments equal to $x$. Fix a compact $K\subset U$ and multi-indices $\beta,\alpha'\in\mathbb{N}_0^n$. Since $\xi$-differentiation commutes with the diagonal evaluation, $\partial_\xi^{\alpha'}c_\alpha(x,\xi)=\frac{(-i)^{|\alpha|}}{\alpha!}[\partial_\xi^{\alpha+\alpha'}\partial_y^\alpha A_0](x,x,\xi)$. Differentiating the diagonal restriction in $x$ distributes the derivative between the first and second slots by the chain rule: \begin{align*} \partial_x^\beta\Big([\partial_\xi^{\alpha+\alpha'}\partial_y^\alpha A_0](x,x,\xi)\Big)=\sum_{\beta'+\beta''=\beta}\binom{\beta}{\beta'}\big[\partial_x^{\beta'}\partial_y^{\beta''+\alpha}\partial_\xi^{\alpha+\alpha'}A_0\big](x,x,\xi). \end{align*} Choose a compact $K_0\subset U$ with $K\times K\subset\operatorname{supp}\chi\cap(K_0\times K_0)$ (possible since $\operatorname{supp}\chi$ contains a neighbourhood of the diagonal of $K$). Applying the order-$m$ amplitude estimate for $A_0$ on $K_0\times K_0\times\mathbb{R}^n$ with $\xi$-multi-index $\alpha+\alpha'$, \begin{align*} \big|\big[\partial_x^{\beta'}\partial_y^{\beta''+\alpha}\partial_\xi^{\alpha+\alpha'}A_0\big](x,x,\xi)\big|\le C\,\langle\xi\rangle^{m-|\alpha+\alpha'|}=C\,\langle\xi\rangle^{m-j-|\alpha'|},\qquad x\in K, \end{align*} using $|\alpha|=j$. Summing the finitely many terms gives $|\partial_x^\beta\partial_\xi^{\alpha'}c_\alpha(x,\xi)|\le C_{K,\alpha',\beta}\,\langle\xi\rangle^{(m-j)-|\alpha'|}$ on $K\times\mathbb{R}^n$, which is exactly the defining estimate for $c_\alpha\in S^{m-j}(U\times\mathbb{R}^n)$. Hence $b_j\in S^{m-j}(U\times\mathbb{R}^n)$ for every $j\in\mathbb{N}_0$. [/step] [step:Asymptotically sum the coefficients into a single symbol $a\in S^m$] [claim:Borel-type summation of the symbol series] Given $b_j\in S^{m-j}(U\times\mathbb{R}^n)$ for $j\in\mathbb{N}_0$, there exists $a\in S^m(U\times\mathbb{R}^n)$ such that $a-\sum_{j=0}^{N-1}b_j\in S^{m-N}(U\times\mathbb{R}^n)$ for every $N\in\mathbb{N}$. [/claim] [proof] Fix $\psi\in C^\infty(\mathbb{R}^n)$ with $0\le\psi\le1$, $\psi(\xi)=0$ for $|\xi|\le1$ and $\psi(\xi)=1$ for $|\xi|\ge2$. Exhaust $U$ by compacts $K_1\subset K_2\subset\cdots$ with $K_i\subset\operatorname{int}K_{i+1}$ and $\bigcup_i K_i=U$. For a symbol $c\in S^\ell$, compact $K$, and integer $\nu\ge0$, write the seminorm \begin{align*} \|c\|_{\ell,K,\nu}=\sup\Big\{\langle\xi\rangle^{-\ell+|\alpha|}\big|\partial_x^\beta\partial_\xi^\alpha c(x,\xi)\big| : x\in K,\ \xi\in\mathbb{R}^n,\ |\alpha|+|\beta|\le\nu\Big\}<\infty. \end{align*} Set $t_0=2$. For $j\ge1$ we choose $t_j\ge\max(2,\,t_{j-1}+1)$ so that \begin{align*} \big\|\psi(\cdot/t_j)\,b_j\big\|_{\,m-j+1,\ K_j,\ j}\le 2^{-j}. \tag{$\ast$} \end{align*} Such a $t_j$ exists: by the Leibniz rule, \begin{align*} \partial_x^\beta\partial_\xi^\alpha\big[\psi(\xi/t_j)b_j(x,\xi)\big]=\sum_{\alpha'\le\alpha}\binom{\alpha}{\alpha'}t_j^{-|\alpha'|}(\partial^{\alpha'}\psi)(\xi/t_j)\,\partial_x^\beta\partial_\xi^{\alpha-\alpha'}b_j(x,\xi). \end{align*} On the support of $\psi(\cdot/t_j)$ we have $|\xi|\ge t_j$, hence $\langle\xi\rangle\ge t_j$. Using $b_j\in S^{m-j}$, the term with $\alpha'=0$ is bounded on $K_j$ by $C\langle\xi\rangle^{m-j-|\alpha|}$, so after multiplying by $\langle\xi\rangle^{-(m-j+1)+|\alpha|}$ it is at most $C\langle\xi\rangle^{-1}\le C\,t_j^{-1}$. For $\alpha'\ne0$, $(\partial^{\alpha'}\psi)(\xi/t_j)$ is supported where $t_j\le|\xi|\le2t_j$, so $\langle\xi\rangle\le 3t_j$ there; the corresponding term, multiplied by the same weight, is bounded by $C\,t_j^{-|\alpha'|}\langle\xi\rangle^{|\alpha'|-1}\le C\,3^{|\alpha'|-1}t_j^{-1}$. Summing the finitely many terms with $|\alpha|+|\beta|\le j$ gives a bound $C_j\,t_j^{-1}$, where $C_j$ depends only on $j$, $\psi$, and the $S^{m-j}$-seminorms of $b_j$ over $K_j$; choosing $t_j$ with $C_j t_j^{-1}\le 2^{-j}$ yields $(\ast)$. Define the map $a:U\times\mathbb{R}^n\to\mathbb{C}$, \begin{align*} a(x,\xi)=\sum_{j=0}^\infty \psi(\xi/t_j)\,b_j(x,\xi). \end{align*} For fixed $\xi$, the summand is nonzero only when $|\xi/t_j|\ge1$, i.e. $t_j\le|\xi|$; since $t_j\to\infty$, only finitely many $j$ contribute, so the sum is locally finite and $a\in C^\infty(U\times\mathbb{R}^n)$. To see $a\in S^m$, fix a compact $K$ and order $\nu$; pick $J\ge\nu$ with $K\subset K_J$. Split $a=\sum_{j<J}\psi(\cdot/t_j)b_j+\sum_{j\ge J}\psi(\cdot/t_j)b_j$. Each term of the finite sum lies in $S^m$ because $b_j\in S^{m-j}\subset S^m$ and $\psi(\cdot/t_j)$ is bounded with bounded derivatives, so the finite sum has finite $S^m$-seminorms over $K$. For the tail, every $j\ge J$ satisfies $j\ge\nu$ and $K\subset K_j$, so $(\ast)$ applies: for $|\alpha|+|\beta|\le\nu$, \begin{align*} \big|\partial_x^\beta\partial_\xi^\alpha[\psi(\xi/t_j)b_j]\big|\le 2^{-j}\langle\xi\rangle^{m-j+1-|\alpha|}\le 2^{-j}\langle\xi\rangle^{m-|\alpha|},\qquad x\in K, \end{align*} using $-j+1\le0$. Summing over $j\ge J$ gives a bound $\le\langle\xi\rangle^{m-|\alpha|}$, so $a\in S^m(U\times\mathbb{R}^n)$. Finally, fix $N$ and write \begin{align*} a-\sum_{j=0}^{N-1}b_j=\sum_{j=0}^{N-1}\big(\psi(\xi/t_j)-1\big)b_j+\sum_{j=N}^{\infty}\psi(\xi/t_j)\,b_j. \end{align*} In the first (finite) sum, $\psi(\xi/t_j)-1$ is smooth and vanishes for $|\xi|\ge2t_j$, hence each $(\psi(\cdot/t_j)-1)b_j$ is smooth with $\xi$-support in the compact set $\{|\xi|\le2t_j\}$; such a function lies in $S^{-\infty}\subset S^{m-N}$. For the tail, fix compact $K$ and order $\nu$ and choose $J'\ge\max(N+1,\nu)$ with $K\subset K_{J'}$. The terms with $N\le j<J'$ are finitely many and each lies in $S^{m-j}\subset S^{m-N}$. For $j\ge J'$, condition $(\ast)$ applies and $-j+1\le -N$, giving for $|\alpha|+|\beta|\le\nu$ \begin{align*} \big|\partial_x^\beta\partial_\xi^\alpha[\psi(\xi/t_j)b_j]\big|\le 2^{-j}\langle\xi\rangle^{m-j+1-|\alpha|}\le 2^{-j}\langle\xi\rangle^{m-N-|\alpha|},\qquad x\in K, \end{align*} and summation over $j\ge J'$ is bounded by $\langle\xi\rangle^{m-N-|\alpha|}$. Hence $a-\sum_{j<N}b_j\in S^{m-N}(U\times\mathbb{R}^n)$. [/proof] Applying the claim to the coefficients $b_j\in S^{m-j}$ from Step 5 produces a symbol $a\in S^m(U\times\mathbb{R}^n)$ with $a-\sum_{j=0}^{N-1}b_j\in S^{m-N}(U\times\mathbb{R}^n)$ for every $N$; this is exactly the asserted asymptotic expansion $a\sim\sum_\alpha\frac1{\alpha!}\partial_\xi^\alpha D_y^\alpha A(x,y,\xi)|_{y=x}$. [/step] [step:Combine the estimates to conclude the difference is smoothing] Let $a$ be the symbol from Step 6 and let $k\in\mathbb{N}_0$ be arbitrary. Choose $N\in\mathbb{N}$ with $N>m+n+k$, so that $m-N<-n-k$. Using $I_A=I_{(1-\chi)A}+I_{A_0}$ (Step 2) and the decomposition of $I_{A_0}$ from Step 4, \begin{align*} I_A-\operatorname{Op}_L(a)=I_{(1-\chi)A}+\Big(\operatorname{Op}_L\big(\textstyle\sum_{j<N}b_j\big)+I_{B_N}\Big)-\operatorname{Op}_L(a) =I_{(1-\chi)A}+I_{B_N}-\operatorname{Op}_L(\rho_N), \end{align*} where $\rho_N:=a-\sum_{j<N}b_j\in S^{m-N}(U\times\mathbb{R}^n)$. The three operators on the right have the following kernels: - $K_{(1-\chi)A}\in C^\infty(U\times U)$ by the claim in Step 2, hence in particular $C^k$; - $K_{B_N}\in C^k(U\times U)$ by the Step 1 claim, since $B_N$ is an amplitude of order $m-N<-n-k$; - the kernel of $\operatorname{Op}_L(\rho_N)$ is $K_{\rho_N}$, where $\rho_N(x,\xi)$, regarded as an amplitude independent of $y$, has order $m-N<-n-k$; the Step 1 claim gives $K_{\rho_N}\in C^k(U\times U)$. Therefore the Schwartz kernel of $I_A-\operatorname{Op}_L(a)$ equals $K_{(1-\chi)A}+K_{B_N}-K_{\rho_N}\in C^k(U\times U)$. The operator $I_A-\operatorname{Op}_L(a)$ does not depend on $N$, so its kernel is a single distribution that coincides with a $C^k$ function for every $k\in\mathbb{N}_0$; hence that kernel belongs to $C^\infty(U\times U)$. This proves that $I_A-\operatorname{Op}_L(a):C_c^\infty(U)\to C^\infty(U)$ is smoothing, with $a\in S^m(U\times\mathbb{R}^n)$, $b_j\in S^{m-j}(U\times\mathbb{R}^n)$, and $a-\sum_{j=0}^{N-1}b_j\in S^{m-N}(U\times\mathbb{R}^n)$ for all $N$, as claimed. [/step]