[proofplan] The proof reduces the Sobolev estimate to an $L^2$ estimate by conjugating $A_h$ with the semiclassical order-reduction operators $\Lambda_h^r=\operatorname{Op}_h(\langle \xi\rangle^r)$. The conjugated operator $\Lambda_h^{s-m}A_h\Lambda_h^{-s}$ has order $0$ by the semiclassical composition theorem, with symbol seminorms controlled by finitely many seminorms of $a$. The semiclassical Calderon-Vaillancourt theorem then gives a uniform $L^2$ bound, which is exactly the desired $H_h^s\to H_h^{s-m}$ bound after rewriting the norms. Finally, density of $\mathcal{S}(\mathbb{R}^n)$ gives the bounded extension to all of $H_h^s(\mathbb{R}^n)$. [/proofplan]

[step:Introduce the semiclassical order-reduction operators] For each $r\in\mathbb{R}$, define the symbol \begin{align*} \lambda_r:\mathbb{R}^n\times\mathbb{R}^n\to(0,\infty),\qquad \lambda_r(x,\xi):=\langle \xi\rangle^r. \end{align*} Let \begin{align*} \Lambda_h^r := \operatorname{Op}_h(\lambda_r):\mathcal{S}(\mathbb{R}^n)\to\mathcal{S}(\mathbb{R}^n). \end{align*} Since $\lambda_r$ is independent of $x$, $\Lambda_h^r$ is the Fourier multiplier $\langle hD\rangle^r$. We use the equivalent defining norm \begin{align*} \|u\|_{H_h^r(\mathbb{R}^n)} := \|\Lambda_h^r u\|_{L^2(\mathbb{R}^n)} \end{align*} for $u\in\mathcal{S}(\mathbb{R}^n)$, and then by completion for $H_h^r(\mathbb{R}^n)$. The Fourier multiplier identity gives \begin{align*} \Lambda_h^r\Lambda_h^{-r}=\operatorname{id}_{\mathcal{S}(\mathbb{R}^n)}. \end{align*} [/step]

[step:Conjugate the operator to order zero]Define \begin{align*} B_h:\mathcal{S}(\mathbb{R}^n)\to\mathcal{S}'(\mathbb{R}^n),\qquad B_h:=\Lambda_h^{s-m}A_h\Lambda_h^{-s}. \end{align*} The symbols $\lambda_{s-m}$ and $\lambda_{-s}$ belong to $S^{s-m}(T^*\mathbb{R}^n)$ and $S^{-s}(T^*\mathbb{R}^n)$, respectively, with seminorms depending only on $s,m,n$ and the chosen multi-index bounds. Applying the semiclassical composition theorem for pseudodifferential operators (citing a result not yet in the wiki: semiclassical pseudodifferential composition theorem), the operator $B_h$ is a semiclassical pseudodifferential operator of order \begin{align*} (s-m)+m+(-s)=0. \end{align*} Thus there exists a symbol family $b=b(x,\xi;h)\in S^0(T^*\mathbb{R}^n)$ such that \begin{align*} B_h=\operatorname{Op}_h(b). \end{align*} Moreover, for each pair of multi-indices $\alpha,\beta$, there are an integer $N_{\alpha,\beta}=N_{\alpha,\beta}(n,m,s)$ and a constant $C_{\alpha,\beta}=C_{\alpha,\beta}(n,m,s,h_0)>0$ such that \begin{align*} q_{\alpha,\beta}^{(0)}(b)\leq C_{\alpha,\beta}\max_{|\gamma|+|\delta|\leq N_{\alpha,\beta}}q_{\gamma,\delta}^{(m)}(a). \end{align*}[/step]

[guided]The purpose of this conjugation is to remove the Sobolev weights from the estimate. The operator $A_h$ has order $m$, so it should lose $m$ derivatives. Multiplying on the left by $\Lambda_h^{s-m}$ measures the output in $H_h^{s-m}$, while multiplying on the right by $\Lambda_h^{-s}$ converts an arbitrary $L^2$ input into an $H_h^s$ input. We define \begin{align*} B_h:\mathcal{S}(\mathbb{R}^n)\to\mathcal{S}'(\mathbb{R}^n),\qquad B_h:=\Lambda_h^{s-m}A_h\Lambda_h^{-s}. \end{align*} The three factors have symbolic orders $s-m$, $m$, and $-s$. The symbols of the two order-reduction factors are $\lambda_{s-m}(x,\xi)=\langle \xi\rangle^{s-m}$ and $\lambda_{-s}(x,\xi)=\langle \xi\rangle^{-s}$. These symbols are fixed once $s,m,n$ are fixed; their seminorms do not depend on $h$ or on the symbol $a$. We now apply the semiclassical composition theorem for pseudodifferential operators (citing a result not yet in the wiki: semiclassical pseudodifferential composition theorem). Its hypotheses are satisfied because $a\in S^m(T^*\mathbb{R}^n)$ uniformly for $0<h\leq h_0$, while $\lambda_{s-m}\in S^{s-m}(T^*\mathbb{R}^n)$ and $\lambda_{-s}\in S^{-s}(T^*\mathbb{R}^n)$. The theorem says that the composition is again a semiclassical pseudodifferential operator, and that its order is the sum of the orders: \begin{align*} (s-m)+m+(-s)=0. \end{align*} Hence there is a symbol family $b=b(x,\xi;h)\in S^0(T^*\mathbb{R}^n)$ such that \begin{align*} B_h=\operatorname{Op}_h(b). \end{align*} The same composition theorem also gives the required seminorm control. For each $\alpha,\beta$, only finitely many derivatives of $a$, $\lambda_{s-m}$, and $\lambda_{-s}$ enter the expansion and the remainder estimate. Since the seminorms of $\lambda_{s-m}$ and $\lambda_{-s}$ are fixed constants depending only on $s,m,n$, these constants can be absorbed into $C_{\alpha,\beta}$. Therefore there are $N_{\alpha,\beta}=N_{\alpha,\beta}(n,m,s)$ and $C_{\alpha,\beta}=C_{\alpha,\beta}(n,m,s,h_0)>0$ such that \begin{align*} q_{\alpha,\beta}^{(0)}(b)\leq C_{\alpha,\beta}\max_{|\gamma|+|\delta|\leq N_{\alpha,\beta}}q_{\gamma,\delta}^{(m)}(a). \end{align*}[/guided]

[step:Apply the uniform semiclassical Calderon-Vaillancourt estimate] By the semiclassical Calderon-Vaillancourt theorem (citing a result not yet in the wiki: semiclassical Calderon-Vaillancourt theorem), there exist an integer $M=M(n)$ and a constant $C_{\mathrm{CV}}=C_{\mathrm{CV}}(n)>0$ such that every order-zero semiclassical operator $\operatorname{Op}_h(b)$ satisfies \begin{align*} \|\operatorname{Op}_h(b)v\|_{L^2(\mathbb{R}^n)}\leq C_{\mathrm{CV}}\left(\max_{|\alpha|+|\beta|\leq M}q_{\alpha,\beta}^{(0)}(b)\right)\|v\|_{L^2(\mathbb{R}^n)} \end{align*} for all $v\in\mathcal{S}(\mathbb{R}^n)$ and all $0<h\leq h_0$. Applying this to $B_h=\operatorname{Op}_h(b)$ and using the seminorm estimate from the previous step, choose \begin{align*} N:=\max_{|\alpha|+|\beta|\leq M}N_{\alpha,\beta} \end{align*} and \begin{align*} C_0:=C_{\mathrm{CV}}\max_{|\alpha|+|\beta|\leq M}C_{\alpha,\beta}. \end{align*} Then for every $v\in\mathcal{S}(\mathbb{R}^n)$, \begin{align*} \|B_hv\|_{L^2(\mathbb{R}^n)}\leq C_0\left(\max_{|\gamma|+|\delta|\leq N}q_{\gamma,\delta}^{(m)}(a)\right)\|v\|_{L^2(\mathbb{R}^n)}. \end{align*} [/step]

[step:Rewrite the $L^2$ bound as the Sobolev mapping estimate] Let $u\in\mathcal{S}(\mathbb{R}^n)$ and define \begin{align*} v:\mathbb{R}^n\to\mathbb{C},\qquad v:=\Lambda_h^s u. \end{align*} Since $\Lambda_h^{-s}\Lambda_h^s=\operatorname{id}_{\mathcal{S}(\mathbb{R}^n)}$, we have \begin{align*} B_hv=\Lambda_h^{s-m}A_h\Lambda_h^{-s}\Lambda_h^s u=\Lambda_h^{s-m}A_hu. \end{align*} Using the $L^2$ bound for $B_h$ gives \begin{align*} \|\Lambda_h^{s-m}A_hu\|_{L^2(\mathbb{R}^n)}\leq C_0\left(\max_{|\gamma|+|\delta|\leq N}q_{\gamma,\delta}^{(m)}(a)\right)\|\Lambda_h^s u\|_{L^2(\mathbb{R}^n)}. \end{align*} By the definition of the semiclassical Sobolev norms, this is exactly \begin{align*} \|A_hu\|_{H_h^{s-m}(\mathbb{R}^n)}\leq C_0\left(\max_{|\gamma|+|\delta|\leq N}q_{\gamma,\delta}^{(m)}(a)\right)\|u\|_{H_h^s(\mathbb{R}^n)}. \end{align*} [/step]

[step:Extend the estimate from Schwartz functions to $H_h^s(\mathbb{R}^n)$] The density theorem for Schwartz functions in semiclassical Sobolev spaces (citing a result not yet in the wiki: density of $\mathcal{S}(\mathbb{R}^n)$ in $H_h^s(\mathbb{R}^n)$) implies that $\mathcal{S}(\mathbb{R}^n)$ is dense in $H_h^s(\mathbb{R}^n)$. The estimate from the previous step shows that $A_h$ is uniformly bounded on this dense subspace as a map into $H_h^{s-m}(\mathbb{R}^n)$. Therefore $A_h$ extends uniquely by continuity to a bounded linear operator \begin{align*} A_h:H_h^s(\mathbb{R}^n)\to H_h^{s-m}(\mathbb{R}^n). \end{align*} The same inequality, with the same constant $C_0$ and the same finite family of symbol seminorms, holds for the extension. This proves the asserted uniform semiclassical Sobolev mapping estimate. [/step]