[proofplan] The bilinear transpose is obtained by interchanging the two spatial variables in the Schwartz kernel and, because the transpose is bilinear rather than Hermitian, no complex conjugation appears. After the change of frequency variable $\eta=-\xi$, the transposed kernel has the usual Kohn-Nirenberg phase but with $x$-$y$ amplitude $a(y,-\xi)$. We then reduce this amplitude to a left Kohn-Nirenberg symbol by Taylor expanding in the spatial input variable around the output variable and integrating by parts in $\xi$. The zeroth term gives $a(x,-\xi)$, and the higher Taylor terms give the stated asymptotic correction series. [/proofplan] [step:Identify the bilinear transpose kernel by interchanging the spatial variables] Let $K_A \in \mathcal S'(\mathbb R^n_x \times \mathbb R^n_y)$ denote the Schwartz kernel of $A$. In oscillatory form, \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*} This identity is understood in the standard oscillatory integral sense. For $u,v \in \mathcal S(\mathbb R^n)$, the bilinear pairing defining the transpose gives \begin{align*} ({}^tA u)(v) = (A v)(u). \end{align*} Writing the right-hand side through the kernel $K_A$ gives \begin{align*} (A v)(u) = \int_{\mathbb R^n}\int_{\mathbb R^n} K_A(x,y)v(y)u(x)\, d\mathcal L^n(y)\, d\mathcal L^n(x). \end{align*} Renaming the variables $x$ and $y$, the Schwartz kernel of ${}^tA$ is therefore \begin{align*} K_{{}^tA}(x,y) = K_A(y,x). \end{align*} Consequently, \begin{align*} K_{{}^tA}(x,y) = (2\pi)^{-n}\int_{\mathbb R^n} e^{i(y-x)\cdot \eta}a(y,\eta)\, d\mathcal L^n(\eta). \end{align*} No complex conjugation has entered the calculation, because ${}^tA$ is the bilinear distributional transpose, not the Hilbert-space adjoint. [/step] [step:Rewrite the transposed kernel with the standard Kohn-Nirenberg phase] Define the amplitude \begin{align*} b: \mathbb R^n_x \times \mathbb R^n_y \times \mathbb R^n_\xi \to \mathbb C \end{align*} by \begin{align*} b(x,y,\xi) = a(y,-\xi). \end{align*} Apply the change of variables $\eta=-\xi$ in the oscillatory integral for $K_{{}^tA}$. The map $\xi \mapsto -\xi$ preserves [Lebesgue measure](/page/Lebesgue%20Measure), so $d\mathcal L^n(\eta)=d\mathcal L^n(\xi)$, and \begin{align*} e^{i(y-x)\cdot \eta} = e^{i(x-y)\cdot \xi}. \end{align*} Thus \begin{align*} K_{{}^tA}(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*} Since $a \in S^m_{1,0}$ and the map $\xi \mapsto -\xi$ preserves the estimates defining $S^m_{1,0}$, the amplitude $b$ satisfies the same order-$m$ symbol estimates in $\xi$, uniformly after differentiating in $x$ and $y$. Here $b$ is independent of $x$, so $x$-derivatives of $b$ vanish. [/step] [step:Reduce the amplitude $a(y,-\xi)$ to a left symbol] For a multi-index $\alpha \in \mathbb N_0^n$, let $D_y^\alpha=(1/i)^{|\alpha|}\partial_y^\alpha$. The amplitude $b$ belongs to the standard Kohn-Nirenberg amplitude class $S^m_{1,0}(\mathbb R^n_x \times \mathbb R^n_y \times \mathbb R^n_\xi)$: for all multi-indices $\beta,\gamma,\delta \in \mathbb N_0^n$ there is a constant $C_{\beta\gamma\delta}>0$ such that \begin{align*} |\partial_x^\beta\partial_y^\gamma\partial_\xi^\delta b(x,y,\xi)| \leq C_{\beta\gamma\delta}(1+|\xi|)^{m-|\delta|} \end{align*} for all $x,y,\xi \in \mathbb R^n$. Indeed, $\partial_x^\beta b=0$ when $|\beta|>0$, and when $\beta=0$ the estimate is exactly the defining symbol estimate for $a(y,-\xi)$, with the sign change in $\xi$ absorbed by the chain rule. For such amplitudes, the Kohn-Nirenberg amplitude-to-left-symbol reduction gives a symbol $c \in S^m_{1,0}(\mathbb R^n_x \times \mathbb R^n_\xi)$ with \begin{align*} c(x,\xi) \sim \sum_{\alpha \in \mathbb N_0^n} \frac{1}{\alpha!}\,\partial_\xi^\alpha D_y^\alpha b(x,y,\xi)\big|_{y=x}, \end{align*} and the operator with kernel \begin{align*} (2\pi)^{-n}\int_{\mathbb R^n}e^{i(x-y)\cdot \xi}b(x,y,\xi)\, d\mathcal L^n(\xi) \end{align*} differs from $\operatorname{Op}(c)$ by an operator with smoothing kernel. This is the standard Kohn-Nirenberg [amplitude reduction theorem](/theorems/7671) for amplitudes in $S^m_{1,0}(\mathbb R^n_x \times \mathbb R^n_y \times \mathbb R^n_\xi)$; if this result is not yet in the wiki, it is the external prerequisite used here. Applying this reduction to $b(x,y,\xi)=a(y,-\xi)$ yields \begin{align*} D_y^\alpha b(x,y,\xi)\big|_{y=x} = D_x^\alpha a(x,-\xi). \end{align*} Therefore \begin{align*} c(x,\xi) \sim \sum_{\alpha \in \mathbb N_0^n} \frac{1}{\alpha!}\,\partial_\xi^\alpha D_x^\alpha\bigl(a(x,-\xi)\bigr). \end{align*} [guided] We now explain why the amplitude reduction has exactly this sign. For a multi-index $\alpha \in \mathbb N_0^n$, $D_y^\alpha$ means $(1/i)^{|\alpha|}\partial_y^\alpha$. The operator obtained after transposition has the standard phase $e^{i(x-y)\cdot \xi}$ but its amplitude depends on the input variable $y$, namely $b(x,y,\xi)=a(y,-\xi)$. This amplitude lies in $S^m_{1,0}(\mathbb R^n_x \times \mathbb R^n_y \times \mathbb R^n_\xi)$ because every derivative $\partial_x^\beta\partial_y^\gamma\partial_\xi^\delta b$ is either zero when $|\beta|>0$ or is a signed derivative of $a(y,-\xi)$, hence satisfies the estimate $C_{\beta\gamma\delta}(1+|\xi|)^{m-|\delta|}$. A left Kohn-Nirenberg symbol must depend on the output variable $x$ and the frequency variable $\xi$, so the task is to replace the $y$-dependence by an asymptotic expansion at $y=x$. Taylor expansion in the $y$ variable gives, for each integer $N \geq 1$, \begin{align*} b(x,y,\xi) = \sum_{|\alpha|<N}\frac{1}{\alpha!}(y-x)^\alpha \partial_y^\alpha b(x,x,\xi) + r_N(x,y,\xi), \end{align*} where $r_N$ is the Taylor remainder. Since $D_y=(1/i)\partial_y$, the derivative $\partial_y^\alpha b$ equals $i^{|\alpha|}D_y^\alpha b$. The key phase identity is \begin{align*} (y-x)^\alpha e^{i(x-y)\cdot \xi} = (-1)^{|\alpha|}D_\xi^\alpha e^{i(x-y)\cdot \xi}, \end{align*} where $D_\xi=(1/i)\partial_\xi$. Integrating by parts in $\xi$ transfers $D_\xi^\alpha$ from the exponential to the symbol factor and contributes another factor $(-1)^{|\alpha|}$. These two signs cancel. After converting $\partial_y^\alpha$ to $D_y^\alpha$, the resulting left-symbol coefficient is therefore \begin{align*} \frac{1}{\alpha!}\partial_\xi^\alpha D_y^\alpha b(x,y,\xi)\big|_{y=x}. \end{align*} For the present amplitude $b(x,y,\xi)=a(y,-\xi)$, differentiating in $y$ and then setting $y=x$ gives \begin{align*} D_y^\alpha b(x,y,\xi)\big|_{y=x} = D_x^\alpha a(x,-\xi). \end{align*} Thus the reduced left symbol has the expansion \begin{align*} c(x,\xi) \sim \sum_{\alpha \in \mathbb N_0^n} \frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha\bigl(a(x,-\xi)\bigr). \end{align*} The Taylor remainder produces, after choosing $N$ arbitrarily large, a remainder in successively lower symbol classes; asymptotic summation for $S^m_{1,0}$ symbols then packages these coefficients into an actual symbol $c \in S^m_{1,0}$, unique modulo $S^{-\infty}$. [/guided] [/step] [step:Check the equivalent sign convention] For each multi-index $\alpha \in \mathbb N_0^n$, the $x$-derivatives commute with the substitution $\xi \mapsto -\xi$, while each $\xi$-derivative contributes a factor $-1$. Hence \begin{align*} \partial_\xi^\alpha D_x^\alpha\bigl(a(x,-\xi)\bigr) = (-1)^{|\alpha|}(\partial_\xi^\alpha D_x^\alpha a)(x,-\xi). \end{align*} Substituting this identity into the preceding expansion gives the equivalent formula \begin{align*} c(x,\xi) \sim \sum_{\alpha \in \mathbb N_0^n} \frac{(-1)^{|\alpha|}}{\alpha!}(\partial_\xi^\alpha D_x^\alpha a)(x,-\xi). \end{align*} [/step] [step:Define the transpose symbol and read off the principal term] Choose ${}^ta \in S^m_{1,0}(\mathbb R^n_x \times \mathbb R^n_\xi)$ to be any asymptotic sum of the coefficients \begin{align*} \frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha\bigl(a(x,-\xi)\bigr). \end{align*} By the amplitude reduction step, ${}^tA-\operatorname{Op}({}^ta)$ has a smoothing kernel, and different choices of asymptotic sum differ by an element of $S^{-\infty}$. The term with $\alpha=0$ is \begin{align*} a(x,-\xi). \end{align*} Every term with $|\alpha|\geq 1$ lies in $S^{m-|\alpha|}_{1,0}\subset S^{m-1}_{1,0}$. Therefore the principal symbol class of ${}^tA$ in $S^m_{1,0}/S^{m-1}_{1,0}$ is represented by $a(x,-\xi)$, as claimed. [/step]