[proofplan] We use the adjoint symbol formula for the global Kohn-Nirenberg calculus. That formula gives a symbol $a^* \in S^m_{1,0}$ for the formal adjoint and an asymptotic expansion whose leading term is $\overline{a}$. Every other term contains at least one derivative in the covariable $\xi$, so it has order at most $m-1$. Passing to the quotient by $S^{m-1}_{1,0}$ leaves only the class of $\overline{a}$. [/proofplan]

[step:Apply the adjoint symbol formula to represent $A^*$ by a symbol of order $m$]Let \begin{align*} (\cdot,\cdot)_{L^2}: \mathcal{S}(\mathbb{R}^n)\times \mathcal{S}(\mathbb{R}^n) \to \mathbb{C} \end{align*} denote the standard complex $L^2$ pairing, given by \begin{align*} (u,v)_{L^2} = \int_{\mathbb{R}^n} u(x)\overline{v(x)}\,d\mathcal{L}^n(x). \end{align*} The space $\mathcal{S}'(\mathbb{R}^n)$ denotes the tempered-distribution dual of the [Schwartz space](/page/Schwartz%20Space) $\mathcal{S}(\mathbb{R}^n)$. The formal adjoint $A^*: \mathcal{S}(\mathbb{R}^n)\to \mathcal{S}'(\mathbb{R}^n)$ is defined by \begin{align*} (Au,v)_{L^2} = (u,A^*v)_{L^2} \end{align*} for all $u,v \in \mathcal{S}(\mathbb{R}^n)$. By the global Kohn-Nirenberg adjoint symbol formula, applied with order $m \in \mathbb{R}$ and symbol $a \in S^m_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)$ acting on the Schwartz class through Kohn-Nirenberg quantization, there exists a symbol $a^* \in S^m_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)$ such that \begin{align*} A^* = \operatorname{Op}(a^*) \end{align*} as operators on $\mathcal{S}(\mathbb{R}^n)$, and \begin{align*} a^* \sim \sum_{\alpha \in \mathbb{N}_0^n}\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha \overline{a}. \end{align*} Here $\mathbb{N}_0:=\mathbb{N}\cup\{0\}$, $\alpha=(\alpha_1,\dots,\alpha_n)$ is a multi-index in $\mathbb{N}_0^n$, $\alpha! = \alpha_1!\cdots\alpha_n!$, $\partial_\xi^\alpha = \partial_{\xi_1}^{\alpha_1}\cdots\partial_{\xi_n}^{\alpha_n}$, and $D_x^\alpha = (-i)^{|\alpha|}\partial_x^\alpha$ is the Kohn-Nirenberg differential convention. Therefore $A^* \in \Psi^m_{1,0}(\mathbb{R}^n)$.[/step]

[guided]The first point is to use the symbolic-calculus theorem that is designed for formal adjoints in the global Kohn-Nirenberg calculus. The formal adjoint is defined relative to the complex $L^2(\mathbb{R}^n)$ pairing \begin{align*} (u,v)_{L^2} = \int_{\mathbb{R}^n} u(x)\overline{v(x)}\,d\mathcal{L}^n(x). \end{align*} Thus $A^*$ is the operator satisfying \begin{align*} (Au,v)_{L^2} = (u,A^*v)_{L^2} \end{align*} for every pair $u,v \in \mathcal{S}(\mathbb{R}^n)$. We apply the global Kohn-Nirenberg adjoint symbol formula. Its hypotheses are exactly the data in the theorem statement: an order $m \in \mathbb{R}$ symbol $a \in S^m_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)$ and the Kohn-Nirenberg operator $A=\operatorname{Op}(a)$ acting on $\mathcal{S}(\mathbb{R}^n)$. The formula gives a symbol $a^* \in S^m_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)$ such that \begin{align*} A^* = \operatorname{Op}(a^*) \end{align*} as operators on $\mathcal{S}(\mathbb{R}^n)$, and it gives the asymptotic expansion \begin{align*} a^* \sim \sum_{\alpha \in \mathbb{N}_0^n}\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha \overline{a}. \end{align*} Here $\mathbb{N}_0:=\mathbb{N}\cup\{0\}$, $\alpha=(\alpha_1,\dots,\alpha_n)$ is a multi-index in $\mathbb{N}_0^n$, $\alpha!=\alpha_1!\cdots\alpha_n!$, $\partial_\xi^\alpha=\partial_{\xi_1}^{\alpha_1}\cdots\partial_{\xi_n}^{\alpha_n}$, and $D_x^\alpha=(-i)^{|\alpha|}\partial_x^\alpha$ is the Kohn-Nirenberg differential convention. This already proves $A^*\in\Psi^m_{1,0}(\mathbb{R}^n)$, because $A^*$ is represented by the order $m$ Kohn-Nirenberg symbol $a^*$. It remains to identify the class of $a^*$ modulo lower-order symbols. For each multi-index $\alpha \in \mathbb{N}_0^n$, define \begin{align*} b_\alpha: \mathbb{R}^n\times\mathbb{R}^n \to \mathbb{C}, \quad b_\alpha(x,\xi)=\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha \overline{a}(x,\xi). \end{align*} Complex conjugation preserves the symbol estimates defining $S^m_{1,0}$, applying $D_x^\alpha$ differentiates only in the base variable and therefore does not decrease the order in the $(1,0)$ calculus, and applying $\partial_\xi^\alpha$ differentiates $|\alpha|$ times in the covariable and decreases the order by $|\alpha|$. Hence \begin{align*} b_\alpha \in S^{m-|\alpha|}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n). \end{align*} When $|\alpha|\geq 1$, the inclusion of symbol classes gives \begin{align*} b_\alpha \in S^{m-1}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n). \end{align*} The only term not forced into $S^{m-1}_{1,0}$ is the term with $\alpha=0$, and for that term $\partial_\xi^0D_x^0$ is the identity, so \begin{align*} b_0=\overline{a}. \end{align*} By the definition of an asymptotic expansion of symbols, subtracting the first term of the expansion leaves a symbol one order lower. Therefore \begin{align*} a^*-\overline{a} \in S^{m-1}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n). \end{align*} Thus $a^*$ and $\overline{a}$ determine the same element of the principal-symbol quotient \begin{align*} S^m_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)/S^{m-1}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n). \end{align*} Since $\sigma_m(A^*)$ is the class of $a^*$ and $\sigma_m(A)$ is the class of $a$, we obtain \begin{align*} \sigma_m(A^*) = [a^*] = [\overline{a}] = \overline{[a]} = \overline{\sigma_m(A)}. \end{align*} This proves the principal-symbol identity.[/guided]

[step:Show that every non-leading term has order at most $m-1$] For each multi-index $\alpha \in \mathbb{N}_0^n$, define the symbol \begin{align*} b_\alpha: \mathbb{R}^n\times\mathbb{R}^n \to \mathbb{C}, \quad b_\alpha(x,\xi)=\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha \overline{a}(x,\xi). \end{align*} Since $a \in S^m_{1,0}$, complex conjugation preserves the same symbol estimates, $x$-derivatives do not lower the order in the $(1,0)$ calculus, and each $\xi$-derivative lowers the order by one. Hence \begin{align*} b_\alpha \in S^{m-|\alpha|}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n). \end{align*} In particular, if $|\alpha|\geq 1$, then \begin{align*} b_\alpha \in S^{m-1}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n). \end{align*} The term with $\alpha=0$ is \begin{align*} b_0=\overline{a}. \end{align*} [/step]

[step:Pass from the adjoint expansion to the principal symbol quotient] By the meaning of the asymptotic expansion for $a^*$, subtracting the first term gives \begin{align*} a^*-\overline{a} \in S^{m-1}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n). \end{align*} Therefore $a^*$ and $\overline{a}$ define the same class in the quotient \begin{align*} S^m_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)/S^{m-1}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n). \end{align*} Since $\sigma_m(A^*)$ is the class of $a^*$ and $\sigma_m(A)$ is the class of $a$, the quotient equality is \begin{align*} \sigma_m(A^*) = [a^*] = [\overline{a}] = \overline{[a]} = \overline{\sigma_m(A)}. \end{align*} This proves the claimed principal-symbol identity. [/step]