[proofplan] We compute the Schwartz kernel of $A$ and take its conjugate transpose to identify the kernel of the formal adjoint. The resulting kernel has the same Kohn-Nirenberg phase but amplitude $\overline{a(y,\xi)}$, so we Taylor expand this amplitude in the $y$ variable at $x$. Powers of $y-x$ are converted into $\xi$-derivatives of the exponential, and [integration by parts](/theorems/210) transfers those derivatives to the symbol. The finite remainders have successively lower symbolic order; asymptotic summation then produces a symbol $a^*$ with the announced expansion, and the residual operator is smoothing. [/proofplan] [step:Compute the conjugate-transpose kernel of the formal adjoint] Let $K_A \in \mathcal{S}'(\mathbb{R}^n_x \times \mathbb{R}^n_y)$ denote the Schwartz kernel distribution of $A$. Formally, its oscillatory representative is \begin{align*} K_A(x,y) = (2\pi)^{-n}\int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}a(x,\xi)\, d\mathcal{L}^n(\xi). \end{align*} The formal adjoint $A^*$ is defined by the identity \begin{align*} \int_{\mathbb{R}^n} Au(x)\overline{v(x)}\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n} u(y)\overline{A^*v(y)}\, d\mathcal{L}^n(y) \end{align*} for $u,v \in \mathcal{S}(\mathbb{R}^n)$. Hence the kernel of $A^*$ is the conjugate transpose \begin{align*} K_{A^*}(x,y) = \overline{K_A(y,x)}. \end{align*} The oscillatory kernel identity is justified by inserting Schwartz functions $u,v \in \mathcal{S}(\mathbb{R}^n)$, pairing against the regularised oscillatory integrals, and passing to the oscillatory limit in \mathcal{S}' after the usual cut-off regularisation in $\xi$; all pairings are continuous on [Schwartz space](/page/Schwartz%20Space). Substituting the expression for $K_A(y,x)$ gives \begin{align*} K_{A^*}(x,y) = (2\pi)^{-n}\int_{\mathbb{R}^n} e^{i(x-y)\cdot \xi}\overline{a(y,\xi)}\, d\mathcal{L}^n(\xi). \end{align*} Thus the only obstruction to writing $A^*$ directly in Kohn-Nirenberg form is that the amplitude depends on $y$ rather than on $x$. [/step] [step:Taylor expand the adjoint amplitude at the left variable] Fix $N \in \mathbb{N}$. For each multi-index $\alpha \in \mathbb{N}_0^n$, write \begin{align*} \partial_x^\alpha \overline{a}: \mathbb{R}^n_x \times \mathbb{R}^n_\xi \to \mathbb{C} \end{align*} for the $\alpha$-th derivative of $\overline{a}$ in the first variable. Taylor's formula in the $y$ variable at the base point $x$ gives \begin{align*} \overline{a(y,\xi)} = \sum_{|\alpha|<N}\frac{(y-x)^\alpha}{\alpha!}\partial_x^\alpha\overline{a(x,\xi)} + r_N(x,y,\xi), \end{align*} where the integral remainder is \begin{align*} r_N(x,y,\xi) = N\sum_{|\alpha|=N}\frac{(y-x)^\alpha}{\alpha!} \int_0^1(1-t)^{N-1}\partial_x^\alpha\overline{a(x+t(y-x),\xi)}\, d\mathcal{L}^1(t). \end{align*} Since $a \in S^m_{1,0}$, conjugation preserves the same symbol estimates, and therefore every derivative $\partial_x^\alpha\overline{a}$ lies in $S^m_{1,0}$. [guided] We must rewrite the amplitude $\overline{a(y,\xi)}$ as something depending on $x$, because Kohn-Nirenberg quantisation places the symbol at the left variable of the kernel. Fix $N \in \mathbb{N}$. For every multi-index $\alpha \in \mathbb{N}_0^n$, define \begin{align*} \partial_x^\alpha \overline{a}: \mathbb{R}^n_x \times \mathbb{R}^n_\xi \to \mathbb{C} \end{align*} to be the derivative of $\overline{a}$ in the first variable. [Taylor's theorem](/theorems/827) for the function $y \mapsto \overline{a(y,\xi)}$, expanded at $x$, gives \begin{align*} \overline{a(y,\xi)} = \sum_{|\alpha|<N}\frac{(y-x)^\alpha}{\alpha!}\partial_x^\alpha\overline{a(x,\xi)} + r_N(x,y,\xi). \end{align*} Here the remainder is not an unspecified error term; it is the explicit integral remainder \begin{align*} r_N(x,y,\xi) = N\sum_{|\alpha|=N}\frac{(y-x)^\alpha}{\alpha!} \int_0^1(1-t)^{N-1}\partial_x^\alpha\overline{a(x+t(y-x),\xi)}\, d\mathcal{L}^1(t). \end{align*} This formula is the point where the symbolic order bookkeeping starts. Differentiating a symbol in $x$ does not lower its order in the class $S^m_{1,0}$, while differentiating in $\xi$ lowers the order. Since $a \in S^m_{1,0}$ and complex conjugation preserves the defining seminorm bounds, each map $\partial_x^\alpha\overline{a}$ remains in $S^m_{1,0}$. The order drop will appear only after the factors $(y-x)^\alpha$ are converted into $\xi$-derivatives. [/guided] [/step] [step:Convert powers of $y-x$ into $\xi$-derivatives of the symbol] For each multi-index $\alpha \in \mathbb{N}_0^n$, the phase satisfies \begin{align*} (y-x)^\alpha e^{i(x-y)\cdot \xi} = (-1)^{|\alpha|}(x-y)^\alpha e^{i(x-y)\cdot \xi} = (-1)^{|\alpha|}D_\xi^\alpha e^{i(x-y)\cdot \xi}, \end{align*} where $D_\xi^\alpha := (1/i)^{|\alpha|}\partial_\xi^\alpha$, as fixed in the theorem statement. Insert the Taylor polynomial into the kernel. For the term indexed by $\alpha$, [integration by parts](/theorems/2098) in $\xi$ gives \begin{align*} (2\pi)^{-n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot \xi}\frac{(y-x)^\alpha}{\alpha!}\partial_x^\alpha\overline{a(x,\xi)}\, d\mathcal{L}^n(\xi) = (2\pi)^{-n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot \xi}\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha\overline{a(x,\xi)}\, d\mathcal{L}^n(\xi). \end{align*} Indeed, using $D_x^\alpha=(1/i)^{|\alpha|}\partial_x^\alpha$, the signs and powers of $i$ combine as \begin{align*} (-1)^{|\alpha|}(-D_\xi)^\alpha\partial_x^\alpha = D_\xi^\alpha\partial_x^\alpha = \partial_\xi^\alpha D_x^\alpha. \end{align*} Therefore the Taylor polynomial contributes the Kohn-Nirenberg symbol \begin{align*} b_N: \mathbb{R}^n_x \times \mathbb{R}^n_\xi \to \mathbb{C}, \qquad (x,\xi) \mapsto \sum_{|\alpha|<N}\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha\overline{a(x,\xi)}. \end{align*} Because $\partial_\xi^\alpha$ lowers the symbolic order by $|\alpha|$, each summand belongs to $S^{m-|\alpha|}_{1,0}$, and hence $b_N \in S^m_{1,0}$. [/step] [step:Estimate the finite Taylor remainder as a lower-order symbol contribution] Let $R_N: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n)$ be the operator whose Schwartz kernel is represented by \begin{align*} K_{R_N}(x,y) = (2\pi)^{-n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot \xi}r_N(x,y,\xi)\, d\mathcal{L}^n(\xi). \end{align*} For each pair of multi-indices $\beta,\gamma \in \mathbb{N}_0^n$ and each multi-index $\delta \in \mathbb{N}_0^n$, the chain rule applied to the affine map $(x,y) \mapsto x+t(y-x)$ expresses $\partial_x^\beta\partial_y^\gamma\partial_\xi^\delta \partial_x^\alpha\overline{a(x+t(y-x),\xi)}$ as a finite linear combination of derivatives $\partial_x^{\alpha+\rho}\partial_\xi^\delta\overline{a(x+t(y-x),\xi)}$, where $|\rho| \le |\beta|+|\gamma|$ and the coefficients are polynomials in $t$ and $1-t$. Since $t \in [0,1]$, these coefficients are uniformly bounded. The defining estimates for $a \in S^m_{1,0}$ therefore give constants $C_{\alpha\beta\gamma\delta N} > 0$ such that \begin{align*} |\partial_x^\beta\partial_y^\gamma\partial_\xi^\delta \partial_x^\alpha\overline{a(x+t(y-x),\xi)}| \le C_{\alpha\beta\gamma\delta N}\langle \xi\rangle^{m-|\delta|} \end{align*} for all $x,y,\xi \in \mathbb{R}^n$ and $t \in [0,1]$. Thus the Taylor remainder amplitude has the form of a finite sum of factors $(y-x)^\alpha h_\alpha(x,y,\xi)$ with $|\alpha|=N$, where \begin{align*} h_\alpha: \mathbb{R}^n_x \times \mathbb{R}^n_y \times \mathbb{R}^n_\xi \to \mathbb{C}, \qquad (x,y,\xi) \mapsto \alpha!^{-1}N\int_0^1(1-t)^{N-1}\partial_x^\alpha\overline{a(x+t(y-x),\xi)}\, d\mathcal{L}^1(t) \end{align*} denotes the corresponding amplitude, and every $h_\alpha$ satisfies uniform two-space amplitude estimates of order $m$. Apply the finite [amplitude reduction theorem](/theorems/7671) for Kohn-Nirenberg oscillatory kernels to these amplitudes. Its hypotheses are precisely the uniform $x,y,\xi$ estimates just verified. The identity $(y-x)^\alpha e^{i(x-y)\cdot \xi}=(-1)^{|\alpha|}D_\xi^\alpha e^{i(x-y)\cdot \xi}$ permits integration by parts in $\xi$; after the $N$ integrations by parts, each resulting term contains $N$ total $\xi$-derivatives of an order-$m$ amplitude. Hence the reduced left symbol is a map $c_N: \mathbb{R}^n_x \times \mathbb{R}^n_\xi \to \mathbb{C}$ satisfying \begin{align*} c_N \in S^{m-N}_{1,0}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi). \end{align*} Consequently \begin{align*} A^* - \operatorname{Op}(b_N+c_N) \end{align*} is smoothing, meaning that its Schwartz kernel is $C^\infty$ and the operator maps $\mathcal{S}'(\mathbb{R}^n)$ continuously into $C^\infty(\mathbb{R}^n)$. [guided] The finite remainder is the analytic point where the proof must check more than formal algebra. Define $R_N: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n)$ by the oscillatory kernel \begin{align*} K_{R_N}(x,y) = (2\pi)^{-n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot \xi}r_N(x,y,\xi)\, d\mathcal{L}^n(\xi). \end{align*} The theorem we use is the finite amplitude reduction theorem for Kohn-Nirenberg kernels. In the form needed here, it says the following: if $h: \mathbb{R}^n_x \times \mathbb{R}^n_y \times \mathbb{R}^n_\xi \to \mathbb{C}$ is a smooth amplitude satisfying, for all multi-indices $\beta,\gamma,\delta \in \mathbb{N}_0^n$, estimates $|\partial_x^\beta\partial_y^\gamma\partial_\xi^\delta h(x,y,\xi)| \le C_{\beta\gamma\delta}\langle \xi\rangle^{m-|\delta|}$ uniformly in $x,y,\xi$, then the oscillatory kernel with amplitude $h$ equals the kernel of a Kohn-Nirenberg operator with a left symbol in $S^m_{1,0}$ modulo a smoothing kernel. If the amplitude is $(y-x)^\alpha h(x,y,\xi)$ with $h$ satisfying those same order-$m$ estimates, then the reduced left symbol belongs to $S^{m-|\alpha|}_{1,0}$, because the factor $(y-x)^\alpha$ is converted into $|\alpha|$ integrations by parts in $\xi$. We verify its hypotheses. The formula for $r_N$ is a finite sum over $|\alpha|=N$ of \begin{align*} (y-x)^\alpha N\alpha!^{-1}\int_0^1(1-t)^{N-1}\partial_x^\alpha\overline{a(x+t(y-x),\xi)}\, d\mathcal{L}^1(t). \end{align*} Because $a \in S^m_{1,0}$, the derivatives of $\overline a$ satisfy the same symbol estimates. When $\partial_x^\beta\partial_y^\gamma\partial_\xi^\delta$ falls on $\partial_x^\alpha\overline{a(x+t(y-x),\xi)}$, the chain rule produces only finitely many derivatives $\partial_x^{\alpha+\rho}\partial_\xi^\delta\overline a$ with $|\rho| \le |\beta|+|\gamma|$, multiplied by powers of $t$ and $1-t$. Since $0 \le t \le 1$, those coefficients are uniformly bounded, so the integrand is bounded by a constant times $\langle \xi\rangle^{m-|\delta|}$ uniformly in $x,y,\xi,t$. The interval $[0,1]$ has finite $\mathcal{L}^1$-measure, so the same estimate holds after integration in $t$. Thus the non-polynomial part of the amplitude satisfies the required uniform two-space amplitude estimates of order $m$. Now the polynomial factor is converted into derivatives of the phase. Since $|\alpha|=N$, \begin{align*} (y-x)^\alpha e^{i(x-y)\cdot \xi} = (-1)^N D_\xi^\alpha e^{i(x-y)\cdot \xi}. \end{align*} Integrating by parts in $\xi$ transfers these $N$ derivatives to the order-$m$ amplitude. In the symbol class $S^m_{1,0}$, each $\xi$-derivative lowers order by one, so the reduced left symbol belongs to $S^{m-N}_{1,0}$. Therefore there is a map $c_N: \mathbb{R}^n_x \times \mathbb{R}^n_\xi \to \mathbb{C}$ with \begin{align*} c_N \in S^{m-N}_{1,0}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi) \end{align*} such that $A^* - \operatorname{Op}(b_N+c_N)$ has a smooth Schwartz kernel. This is exactly the finite-order adjoint expansion needed for the asymptotic argument. [/guided] [/step] [step:Use asymptotic summation to construct the adjoint symbol] Define, for every multi-index $\alpha \in \mathbb{N}_0^n$, \begin{align*} q_\alpha(x,\xi) := \frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha\overline{a(x,\xi)}, \qquad q_\alpha: \mathbb{R}^n_x \times \mathbb{R}^n_\xi \to \mathbb{C}. \end{align*} Then $q_\alpha \in S^{m-|\alpha|}_{1,0}$. By asymptotic summation for symbols, applied to the sequence of order bounds $m-|\alpha|$, there exists a map $a^*: \mathbb{R}^n_x \times \mathbb{R}^n_\xi \to \mathbb{C}$ with $a^* \in S^m_{1,0}$ such that, for each $N \in \mathbb{N}$, \begin{align*} a^* - \sum_{|\alpha|<N}q_\alpha \in S^{m-N}_{1,0}. \end{align*} This is precisely the asymptotic expansion \begin{align*} a^*(x,\xi)\sim \sum_{\alpha \in \mathbb{N}_0^n}\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha\overline{a(x,\xi)}. \end{align*} Here $S^{-\infty}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi) := \bigcap_{M \in \mathbb{R}} S^M_{1,0}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi)$. The symbol $a^*$ is determined only modulo $S^{-\infty}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi)$, which is exactly the ambiguity allowed by smoothing operators. [/step] [step:Identify the smoothing residual and extract the principal consequence] Let $S^{-\infty}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi) := \bigcap_{M \in \mathbb{R}}S^M_{1,0}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi)$ denote the residual symbol class. The finite amplitude reduction theorem, applied for each $N \in \mathbb{N}$, says that the full left symbol of the adjoint kernel has partial expansion $b_N$ with remainder in $S^{m-N}_{1,0}$ modulo $S^{-\infty}$. The asymptotic summation theorem constructs $a^*$ with the same partial expansions, namely $a^*-b_N \in S^{m-N}_{1,0}$ for every $N$. Therefore the difference between the reduced left symbol of $A^*$ and $a^*$ lies in $S^{m-N}_{1,0}$ for every $N$, hence lies in $S^{-\infty}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi)$. The smoothing kernel criterion for Kohn-Nirenberg operators states that a symbol in $S^{-\infty}$ has a $C^\infty$ Schwartz kernel and maps $\mathcal{S}'(\mathbb{R}^n)$ continuously into $C^\infty(\mathbb{R}^n)$. It follows that $A^*-\operatorname{Op}(a^*)$ is smoothing. Finally take $N=1$ in the asymptotic expansion. The only multi-index with $|\alpha|<1$ is $\alpha=0$, so \begin{align*} \sum_{|\alpha|<1}\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha\overline{a(x,\xi)} = \overline{a(x,\xi)}. \end{align*} Hence \begin{align*} a^*-\overline{a}\in S^{m-1}_{1,0}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi), \end{align*} which proves the stated principal-symbol consequence and completes the proof. [/step]