[proofplan] We write the localized composition as an oscillatory integral with one intermediate spatial variable and then factor out the left-quantized phase $e^{i(x-y)\cdot \xi}$. This identifies the local composition symbol as an oscillatory integral in the difference variable $z-x$ and the auxiliary frequency variable. Taylor expansion of the localized $b$-factor in the intermediate spatial variable produces the finite symbolic terms, while [integration by parts](/theorems/210) transfers powers of $z-x$ to $\xi$-derivatives of $a$ with the convention $D_x^\alpha=(-i)^{|\alpha|}\partial_x^\alpha$. The Taylor remainder is estimated by the standard compactly supported oscillatory integral calculus, giving the order drop by $N$, and off-diagonal [integration by parts](/theorems/2098) gives smoothing for cutoff changes. [/proofplan]

[step:Write the localized composition kernel with a compact intermediate variable]Let $u \in C_c^\infty(U)$. For $x \in V$, the equality $\chi(x)=1$ holds because $\chi=1$ on an open neighbourhood of $\overline V$. Define the oscillatory kernel $K: V \times U \to \mathbb{C}$ of the localized composition by the following compactly supported oscillatory integral: \begin{align*} K(x,y) = (2\pi)^{-2n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot \eta}a(x,\eta)\psi(z)e^{i(z-y)\cdot \xi}b(z,\xi)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\eta)\,d\mathcal{L}^n(\xi). \end{align*} Here $z \in U$ is the intermediate spatial variable, $\eta,\xi \in \mathbb{R}^n$ are frequency variables, and the factor $\psi(z)$ makes the $z$-support compact in $U$. Since $\psi \in C_c^\infty(U)$, all spatial integrations involving $z$ take place in the compact set $\operatorname{supp}\psi \Subset U$. To justify the displayed formula, choose regularizing cutoffs $\varphi_j,\omega_j \in C_c^\infty(\mathbb{R}^n)$ with $\varphi_j \to 1$ and $\omega_j \to 1$ in the standard oscillatory sense. For fixed $j$, the regularized kernels are absolutely integrable in $(z,\eta,\xi)$ because $z$ is compactly supported and the frequency cutoffs are compactly supported, so [Fubini's theorem](/theorems/2961) applies to the regularized composition. The compact $z$-support and the symbol seminorm estimates for $a$ and $b$ give uniform oscillatory bounds, allowing passage to the limit that defines $K$. Put \begin{align*} \zeta = \eta-\xi. \end{align*} The [Lebesgue measure](/page/Lebesgue%20Measure) in frequency space is invariant under this translation, so \begin{align*} d\mathcal{L}^n(\eta)=d\mathcal{L}^n(\zeta). \end{align*} The phase factors combine as \begin{align*} e^{i(x-z)\cdot \eta}e^{i(z-y)\cdot \xi}=e^{i(x-y)\cdot \xi}e^{i(x-z)\cdot \zeta}. \end{align*} Thus on $V$ the kernel may be written in left-quantized form \begin{align*} K(x,y) = (2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}c_0(x,\xi)\,d\mathcal{L}^n(\xi), \end{align*} where the preliminary local symbol \begin{align*} c_0: V \times \mathbb{R}^n \to \mathbb{C} \end{align*} is defined by \begin{align*} c_0(x,\xi) = (2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot \zeta}a(x,\xi+\zeta)\psi(z)b(z,\xi)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\zeta). \end{align*} This is the local symbol attached to the chosen cutoffs. Choose a function \begin{align*} \theta: U \times U \to \mathbb{C} \end{align*} with $\theta \in C_c^\infty(U \times U)$, $\theta(x,z)=1$ on an open neighbourhood of $\{(x,x):x\in \overline V\}$, $\operatorname{supp}_z\theta(x,\cdot)$ contained in the [open set](/page/Open%20Set) where $\psi=1$ for every $x\in V$, and such that $x+t(z-x)\in U$ for all $(x,z)\in\operatorname{supp}\theta$ with $x\in V$ and all $t\in[0,1]$. This last condition is achieved by shrinking the support of $\theta$ around the diagonal, using $\overline V\Subset U$. Splitting $1=\theta(x,z)+(1-\theta(x,z))$, the part containing $1-\theta$ is smoothing on $V$ by repeated integration by parts in $\zeta$, because $|x-z|$ is bounded below on its support over each compact subset of $V$. Hence, modulo a smoothing symbol, we may replace $c_0$ by the diagonal symbol $c_\Delta: V \times \mathbb{R}^n \to \mathbb{C}$ defined by \begin{align*} c_\Delta(x,\xi)=(2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot\zeta}\theta(x,z)a(x,\xi+\zeta)b(z,\xi)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\zeta). \end{align*}[/step]

[guided]Let us track the variables carefully because this is where the left-quantization convention enters. The operator $\operatorname{Op}(b)$ first sends $u$ to \begin{align*} \operatorname{Op}(b)u(z) = (2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_U e^{i(z-y)\cdot \xi}b(z,\xi)u(y)\,d\mathcal{L}^n(y)\,d\mathcal{L}^n(\xi). \end{align*} Multiplication by $\psi$ gives the function $z \mapsto \psi(z)\operatorname{Op}(b)u(z)$. Applying $\operatorname{Op}(a)$ and then restricting to $x \in V$, where $\chi(x)=1$, gives the kernel \begin{align*} K(x,y) = (2\pi)^{-2n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot \eta}a(x,\eta)\psi(z)e^{i(z-y)\cdot \xi}b(z,\xi)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\eta)\,d\mathcal{L}^n(\xi). \end{align*} The compact support of $\psi$ is essential: it ensures that the intermediate spatial variable $z$ remains in a fixed compact subset of $U$, so the local symbol estimates for $a$ and $b$ may be used uniformly on the relevant $x$- and $z$-sets. Now define the auxiliary frequency variable $\zeta=\eta-\xi$. [Translation invariance of Lebesgue measure](/theorems/4911) gives $d\mathcal{L}^n(\eta)=d\mathcal{L}^n(\zeta)$, and the phase becomes \begin{align*} e^{i(x-z)\cdot \eta}e^{i(z-y)\cdot \xi}=e^{i(x-z)\cdot(\xi+\zeta)}e^{i(z-y)\cdot \xi}=e^{i(x-y)\cdot \xi}e^{i(x-z)\cdot \zeta}. \end{align*} The factor $e^{i(x-y)\cdot \xi}$ is exactly the left-quantized phase. Therefore the remaining oscillatory integral defines the local composition symbol \begin{align*} c_0(x,\xi) = (2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot \zeta}a(x,\xi+\zeta)\psi(z)b(z,\xi)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\zeta). \end{align*} This proves that, on $V$, the localized composition has the form of a left-quantized operator with symbol $c_0$, up to the usual interpretation of the oscillatory integral. We next isolate the diagonal contribution. Choose $\theta \in C_c^\infty(U\times U)$ equal to $1$ near $\{(x,x):x\in\overline V\}$, supported in the $z$-variable where $\psi=1$, and small enough around the diagonal that $x+t(z-x)\in U$ whenever $(x,z)\in\operatorname{supp}\theta$, $x\in V$, and $t\in[0,1]$. Such a choice is possible because $\overline V\Subset U$ and $\psi=1$ on an open neighbourhood of $\operatorname{supp}\chi$. On the support of $1-\theta(x,z)$, with $x$ restricted to a compact subset of $V$, the distance $|x-z|$ has a positive lower bound. The identity \begin{align*} \frac{1-i(x-z)\cdot\nabla_\zeta}{1+|x-z|^2}e^{i(x-z)\cdot\zeta}=e^{i(x-z)\cdot\zeta} \end{align*} then permits repeated integration by parts in $\zeta$, producing arbitrary decay in $\zeta$ and hence a smoothing symbol. Thus the symbolic expansion may be proved for the diagonal symbol $c_\Delta$ with $\theta$ inserted and $\psi$ omitted on the support of $\theta$.[/guided]

[step:Taylor expand the localized right symbol around the output point] Fix an integer $N \ge 1$. We Taylor expand $b(z,\xi)$ on the support of $\theta(x,z)$, not the cutoff factor, because the diagonal cutoff $\theta(x,z)$ will remain in every oscillatory integral. By the support condition imposed on $\theta$, the line segment $\{x+t(z-x):0\le t\le 1\}$ is contained in $U$ whenever $(x,z)\in\operatorname{supp}\theta$ and $x\in V$. Taylor's formula with integral remainder in the spatial variable around $x$ therefore gives, for such $(x,z)$, \begin{align*} b(z,\xi)=\sum_{|\alpha|<N}\frac{(z-x)^\alpha}{\alpha!}\partial_x^\alpha b(x,\xi)+\sum_{|\alpha|=N}\frac{N(z-x)^\alpha}{\alpha!}\int_0^1(1-t)^{N-1}\partial_x^\alpha b(x+t(z-x),\xi)\,d\mathcal{L}^1(t). \end{align*} Here $\alpha=(\alpha_1,\dots,\alpha_n)\in\mathbb{N}_0^n$ is a multi-index and $(z-x)^\alpha=\prod_{j=1}^n(z_j-x_j)^{\alpha_j}$. Thus $c_\Delta$ decomposes as \begin{align*} c_\Delta(x,\xi)=\sum_{|\alpha|<N}c_\alpha(x,\xi)+r_N(x,\xi), \end{align*} where $c_\alpha: V \times \mathbb{R}^n \to \mathbb{C}$ is defined by \begin{align*} c_\alpha(x,\xi)=(2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot \zeta}\theta(x,z)a(x,\xi+\zeta)\frac{(z-x)^\alpha}{\alpha!}\partial_x^\alpha b(x,\xi)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\zeta), \end{align*} and $r_N: V \times \mathbb{R}^n \to \mathbb{C}$ is obtained by inserting the integral remainder for $b(z,\xi)$ while retaining the factor $\theta(x,z)$. [/step]

[step:Convert powers of the spatial difference into derivatives of the left symbol] For each component $j \in \{1,\dots,n\}$, \begin{align*} (z_j-x_j)e^{i(x-z)\cdot\zeta}=i\,\partial_{\zeta_j}e^{i(x-z)\cdot\zeta}. \end{align*} Therefore \begin{align*} (z-x)^\alpha e^{i(x-z)\cdot\zeta}=i^{|\alpha|}\partial_\zeta^\alpha e^{i(x-z)\cdot\zeta}. \end{align*} Integrating by parts in the $\zeta$-variable transfers $\partial_\zeta^\alpha$ from the exponential to $a(x,\xi+\zeta)$ and produces the sign $(-1)^{|\alpha|}$. Since $\partial_\zeta^\alpha a(x,\xi+\zeta)=\partial_\xi^\alpha a(x,\xi+\zeta)$, we get \begin{align*} c_\alpha(x,\xi)=\frac{(-i)^{|\alpha|}}{\alpha!}\partial_x^\alpha b(x,\xi)(2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot\zeta}\theta(x,z)\partial_\xi^\alpha a(x,\xi+\zeta)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\zeta). \end{align*} Because $\theta(x,z)=1$ near $z=x$, the localized oscillatory identity \begin{align*} (2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot\zeta}\theta(x,z)q(x,\xi+\zeta)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\zeta)=q(x,\xi) \quad \operatorname{mod} S^{-\infty}(V\times\mathbb{R}^n) \end{align*} holds for symbols $q$: the part where $\theta=1$ is the inverse Fourier representation of the delta distribution at $z=x$, and the complementary part is smoothing by repeated integration by parts in $\zeta$ on a set where $|x-z|$ is bounded below over compact subsets of $V$. Applying this identity with \begin{align*} q: V \times \mathbb{R}^n \to \mathbb{C}, \qquad q(x,\xi)=\partial_\xi^\alpha a(x,\xi), \end{align*} gives, after absorbing the smoothing contribution into the final smoothing remainder, \begin{align*} c_\alpha(x,\xi)=\frac{1}{\alpha!}\partial_\xi^\alpha a(x,\xi)(-i)^{|\alpha|}\partial_x^\alpha b(x,\xi). \end{align*} Since $D_x^\alpha=(-i)^{|\alpha|}\partial_x^\alpha$, this is exactly \begin{align*} c_\alpha(x,\xi)=\frac{1}{\alpha!}\partial_\xi^\alpha a(x,\xi)D_x^\alpha b(x,\xi). \end{align*} [/step]

[step:Estimate each symbolic coefficient in the expected order] Let $\beta,\gamma \in \mathbb{N}_0^n$ be multi-indices. Since $a \in S^m_{1,0}(U \times \mathbb{R}^n)$, every $\xi$-derivative lowers order by its length, so \begin{align*} \partial_\xi^\alpha a \in S^{m-|\alpha|}_{1,0}(U \times \mathbb{R}^n). \end{align*} Since $b \in S^{m'}_{1,0}(U \times \mathbb{R}^n)$, spatial derivatives do not lower the symbolic order, so \begin{align*} D_x^\alpha b \in S^{m'}_{1,0}(U \times \mathbb{R}^n). \end{align*} The product rule for symbols gives \begin{align*} \partial_\xi^\alpha a \, D_x^\alpha b \in S^{m+m'-|\alpha|}_{1,0}(V \times \mathbb{R}^n). \end{align*} Multiplication by the scalar factor $\alpha!^{-1}$ does not change the order. Hence every displayed coefficient belongs to $S^{m+m'-|\alpha|}_{1,0}(V \times \mathbb{R}^n)$. [/step]

[step:Control the Taylor remainder by compactly supported oscillatory estimates]For $|\alpha|=N$, define \begin{align*} R_\alpha: V \times U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by \begin{align*} R_\alpha(x,z,\xi)=\frac{N}{\alpha!}\int_0^1(1-t)^{N-1}\partial_x^\alpha b(x+t(z-x),\xi)\,d\mathcal{L}^1(t). \end{align*} Then the remainder has the form \begin{align*} r_N(x,\xi)=(2\pi)^{-n}\sum_{|\alpha|=N}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_U e^{i(x-z)\cdot\zeta}\theta(x,z)a(x,\xi+\zeta)(z-x)^\alpha R_\alpha(x,z,\xi)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\zeta). \end{align*} For each compact set $K\Subset V$, the set $\{z\in U:(x,z)\in\operatorname{supp}\theta \text{ for some } x\in K\}$ is compact in $U$. Thus the $z$-support of the full amplitude $\theta(x,z)R_\alpha(x,z,\xi)$ is compact uniformly for $x\in K$. Moreover, for every pair of multi-indices $\beta,\gamma$, the symbol estimates for $b$ and the compact support of $\theta$ imply bounds of the form \begin{align*} |\partial_x^\beta\partial_\xi^\gamma(\theta(x,z)R_\alpha(x,z,\xi))|\le C_{\alpha,\beta,\gamma,K,\theta}(1+|\xi|)^{m'-|\gamma|} \end{align*} for $x\in K$, uniformly in $z \in U$. Using $(z-x)^\alpha e^{i(x-z)\cdot\zeta}=i^{|\alpha|}\partial_\zeta^\alpha e^{i(x-z)\cdot\zeta}$ and integrating by parts in $\zeta$ transfers $N$ frequency derivatives to the amplitude $a(x,\xi+\zeta)$, while $\theta(x,z)R_\alpha(x,z,\xi)$ is independent of $\zeta$. For derivatives in $x$ and $\xi$, the differentiated amplitude is a finite sum of products of derivatives of $\theta$, derivatives of $R_\alpha$, and derivatives $\partial_x^{\beta_1}\partial_\xi^{\gamma_1+\delta}a(x,\xi+\zeta)$ with $|\delta|=N$. The symbol estimates give polynomial bounds in $(\xi,\zeta)$), and the compact support in $z$ allows the standard compactly supported oscillatory integral estimate: if an amplitude $A(x,z,\xi,\zeta)$ is compactly supported in $z$ uniformly for $x\in K$ and satisfies symbol bounds of order $M$ in $\xi$ after the oscillatory integration in $(z,\zeta)$, then $(2\pi)^{-n}\operatorname{Os}\!\int e^{i(x-z)\cdot\zeta}A\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(\zeta)$ is a symbol of order $M$ on $K\times\mathbb{R}^n$. Here the $N$ transferred derivatives lower the $a$-order from $m$ to $m-N$, and the factor $\theta R_\alpha$ has order $m'$, so $M=m+m'-N$. Consequently, for every compact $K \Subset V$ and every pair of multi-indices $\beta,\gamma$, there is a constant $C_{N,\beta,\gamma,K}>0$, depending only on finitely many symbol seminorms of $a$ and $b$, finitely many derivatives of $\theta$, and $K$, such that \begin{align*} |\partial_x^\beta\partial_\xi^\gamma r_N(x,\xi)|\le C_{N,\beta,\gamma,K}(1+|\xi|)^{m+m'-N-|\gamma|} \end{align*} for all $x \in K$ and all $\xi \in \mathbb{R}^n$. Hence \begin{align*} r_N \in S^{m+m'-N}_{1,0}(V \times \mathbb{R}^n). \end{align*}[/step]

[guided]The Taylor remainder contains the factor $(z-x)^\alpha$ with $|\alpha|=N$. This factor is the source of the gain of $N$ orders. Define \begin{align*} R_\alpha(x,z,\xi)=\frac{N}{\alpha!}\int_0^1(1-t)^{N-1}\partial_x^\alpha b(x+t(z-x),\xi)\,d\mathcal{L}^1(t). \end{align*} The function $R_\alpha$ itself need not be compactly supported in $z$; the compact support comes from the diagonal cutoff $\theta(x,z)$ retained in the remainder integral. For every compact $K\Subset V$, the projection of $\operatorname{supp}\theta\cap(K\times U)$ to the $z$-variable is a compact subset of $U$. This is the support condition required by the compactly supported oscillatory integral estimate. For each multi-index $\gamma$, differentiating $R_\alpha$ in $\xi$ differentiates only $b$. Since $b\in S^{m'}_{1,0}(U\times\mathbb{R}^n)$, every $\xi$-derivative lowers the order by its length. Derivatives in $x$ may also hit $\theta$ and the affine point $x+t(z-x)$, but on the compact support just described these only change the constant. Thus for every compact $K\Subset V$ and every multi-index pair $\beta,\gamma$, there is a constant $C_{\alpha,\beta,\gamma,K,\theta}>0$ such that \begin{align*} |\partial_x^\beta\partial_\xi^\gamma(\theta(x,z)R_\alpha(x,z,\xi))|\le C_{\alpha,\beta,\gamma,K,\theta}(1+|\xi|)^{m'-|\gamma|}. \end{align*} The derivatives in $x$ are harmless for the symbol order because this is the $(1,0)$ class. Now use the oscillation. The identity \begin{align*} (z-x)^\alpha e^{i(x-z)\cdot\zeta}=i^{|\alpha|}\partial_\zeta^\alpha e^{i(x-z)\cdot\zeta} \end{align*} lets us integrate by parts in $\zeta$. Since $\theta(x,z)R_\alpha(x,z,\xi)$ does not depend on $\zeta$, the transferred derivatives fall on $a(x,\xi+\zeta)$, apart from the harmless frequency cutoffs used in the definition of the oscillatory integral. When the derivatives fall on $a$, they become $\partial_\xi$-derivatives of $a$, and each such derivative lowers the order of $a$ by one. Therefore the $a$-factor has effective order $m-N$, while the $\theta R_\alpha$ factor has order $m'$. Their product has order $m+m'-N$ after the compactly supported oscillatory integration in $(z,\zeta)$. The compactly supported oscillatory integral estimate applies because the $z$-support is compact uniformly for $x\in K$, all $x$- and $\xi$-derivatives of $\theta R_\alpha$ satisfy the displayed symbol bounds, and the transferred derivatives of $a$ satisfy the symbol bounds of order $m-N$. It gives the precise seminorm bound: for every compact $K\Subset V$ and every multi-index pair $\beta,\gamma$, there is a constant $C_{N,\beta,\gamma,K}>0$ such that \begin{align*} |\partial_x^\beta\partial_\xi^\gamma r_N(x,\xi)|\le C_{N,\beta,\gamma,K}(1+|\xi|)^{m+m'-N-|\gamma|}. \end{align*} This is exactly the defining estimate for \begin{align*} r_N \in S^{m+m'-N}_{1,0}(V \times \mathbb{R}^n). \end{align*}[/guided]

[step:Assemble the asymptotic expansion] For each integer $N\ge 1$, the preceding steps give, modulo a smoothing symbol absorbed into $r_N$, \begin{align*} c_0(x,\xi)-\sum_{|\alpha|<N}\frac{1}{\alpha!}\partial_\xi^\alpha a(x,\xi)D_x^\alpha b(x,\xi)=r_N(x,\xi) \end{align*} with \begin{align*} r_N \in S^{m+m'-N}_{1,0}(V \times \mathbb{R}^n). \end{align*} Taking $N=1$ gives $c_0 \in S^{m+m'}_{1,0}(V \times \mathbb{R}^n)$. Since the same remainder estimate holds for every $N$, the definition of asymptotic expansion in the filtered symbol classes gives \begin{align*} c_0(x,\xi)\sim \sum_{\alpha\in\mathbb{N}_0^n}\frac{1}{\alpha!}\partial_\xi^\alpha a(x,\xi)D_x^\alpha b(x,\xi). \end{align*} Set \begin{align*} c: V \times \mathbb{R}^n \to \mathbb{C}, \qquad c(x,\xi)=c_0(x,\xi). \end{align*} Then $\operatorname{Op}(c)$ represents the localized composition on $V$ modulo smoothing terms. [/step]

[step:Show that cutoff changes contribute only smoothing symbols on $V$] Consider two admissible choices of auxiliary cutoffs, and let $\psi_1,\psi_2 \in C_c^\infty(U)$ be their middle cutoffs. Their difference \begin{align*} \rho: U \to \mathbb{C}, \qquad \rho(z)=\psi_1(z)-\psi_2(z) \end{align*} vanishes on an open neighbourhood of $\operatorname{supp}\chi$. In particular, since $x\in V$ and $\chi=1$ on a neighbourhood of $\overline V$, the pairs $(x,z)$ with $z\in\operatorname{supp}\rho$ stay a positive distance away from the diagonal $x=z$ after $x$ is restricted to compact subsets of $V$. The difference of the corresponding kernels is a finite sum of oscillatory kernels whose phase contains $e^{i(x-z)\cdot\zeta}$ with $|x-z|$ bounded below on the support of the amplitude. Define the integration-by-parts operator \begin{align*} L_{x,z,\zeta}: C^\infty(\mathbb{R}^n_\zeta) \to C^\infty(\mathbb{R}^n_\zeta), \qquad L_{x,z,\zeta}=\frac{1-i(x-z)\cdot\nabla_\zeta}{1+|x-z|^2}. \end{align*} Its formal transpose satisfies \begin{align*} L_{x,z,\zeta}^{\,t}e^{i(x-z)\cdot\zeta}=e^{i(x-z)\cdot\zeta}. \end{align*} Repeated integration by parts with $L_{x,z,\zeta}$ gives an arbitrary power of decay in $\zeta$ because $|x-z|$ is bounded below and all $z$-derivatives remain compactly supported. Consequently the cutoff-difference symbol lies in \begin{align*} S^{-\infty}(V \times \mathbb{R}^n)=\bigcap_{M\in\mathbb{R}}S^M_{1,0}(V \times \mathbb{R}^n). \end{align*} The same off-diagonal argument applies to changes of the outer cutoff $\chi$ and to properly supported representatives. Here a properly supported representative of $\operatorname{Op}(a)$ or $\operatorname{Op}(b)$ means an operator whose Schwartz kernel is properly supported over $U\times U$ and agrees with the corresponding local left-quantized kernel in a neighbourhood of the diagonal; the difference kernel is therefore supported away from the relevant diagonal over $V$. Hence the same integration-by-parts argument gives a smoothing symbol. Therefore changing the auxiliary cutoffs changes the localized symbol only by a smoothing symbol, and the corresponding operator changes only by a smoothing operator on $V$. Combining this smoothing independence with the asymptotic expansion proved above completes the proof. [/step]