[proofplan] We conjugate $A$ by Bessel potential operators so that the desired $H^s\to H^{s-m}$ estimate becomes an $L^2\to L^2$ estimate for an order-zero pseudodifferential operator. The type $(1,0)$ composition theorem shows that this conjugated operator has order $0$, with finitely many order-zero symbol seminorms controlled by finitely many order-$m$ seminorms of $a$. Calderón-Vaillancourt then gives the $L^2$ bound. Finally, density of $\mathcal{S}(\mathbb{R}^n)$ in $H^s(\mathbb{R}^n)$ gives the bounded extension, and uniqueness follows from continuity on a dense subspace. [/proofplan]

[step:Introduce the Bessel potential operators and reduce the estimate to $L^2$]For each $t\in \mathbb{R}$, define the Bessel symbol $j_t:\mathbb{R}^n\to (0,\infty)$ by $j_t(\xi)=\langle \xi\rangle^t$, and define the Bessel potential operator $J^t:\mathcal{S}(\mathbb{R}^n)\to \mathcal{S}(\mathbb{R}^n)$ by \begin{align*} \widehat{J^t u}(\xi)=\langle \xi\rangle^t\hat{u}(\xi). \end{align*} The Sobolev norm is \begin{align*} \|u\|_{H^t(\mathbb{R}^n)}=\|J^t u\|_{L^2(\mathbb{R}^n)}. \end{align*} Let $u\in \mathcal{S}(\mathbb{R}^n)$ and set $v:=J^s u\in \mathcal{S}(\mathbb{R}^n)$. Since $J^{-s}J^s u=u$ on $\mathcal{S}(\mathbb{R}^n)$, and since $\mathcal{S}'(\mathbb{R}^n)$ denotes the space of [tempered distributions](/page/Tempered%20Distributions) on $\mathbb{R}^n$, define \begin{align*} B:=J^{s-m}AJ^{-s}:\mathcal{S}(\mathbb{R}^n)\to \mathcal{S}'(\mathbb{R}^n). \end{align*} Then \begin{align*} \|Au\|_{H^{s-m}(\mathbb{R}^n)}=\|J^{s-m}Au\|_{L^2(\mathbb{R}^n)}=\|Bv\|_{L^2(\mathbb{R}^n)}. \end{align*} Thus it is enough to prove an $L^2$ estimate for $B$ in terms of finitely many seminorms of $a$.[/step]

[guided]The purpose of the Bessel operators is to translate Sobolev regularity into an $L^2$ statement. For each $t\in\mathbb{R}$, the symbol $j_t:\mathbb{R}^n\to(0,\infty)$ is defined by \begin{align*} j_t(\xi)=\langle \xi\rangle^t. \end{align*} The operator $J^t:\mathcal{S}(\mathbb{R}^n)\to\mathcal{S}(\mathbb{R}^n)$ is the Fourier multiplier with symbol $j_t$, so \begin{align*} \widehat{J^t u}(\xi)=\langle \xi\rangle^t\hat{u}(\xi). \end{align*} By definition of the $H^t(\mathbb{R}^n)$ norm, \begin{align*} \|u\|_{H^t(\mathbb{R}^n)}=\|J^t u\|_{L^2(\mathbb{R}^n)}. \end{align*} Now take $u\in\mathcal{S}(\mathbb{R}^n)$ and define $v:=J^s u$. Since $J^s$ and $J^{-s}$ are inverse Fourier multipliers on $\mathcal{S}(\mathbb{R}^n)$, we have $J^{-s}v=u$. Therefore \begin{align*} J^{s-m}Au=J^{s-m}AJ^{-s}v. \end{align*} This motivates the definition \begin{align*} B:=J^{s-m}AJ^{-s}. \end{align*} With this notation, \begin{align*} \|Au\|_{H^{s-m}(\mathbb{R}^n)}=\|J^{s-m}Au\|_{L^2(\mathbb{R}^n)}=\|Bv\|_{L^2(\mathbb{R}^n)}. \end{align*} So the Sobolev estimate for $A$ will follow once we prove that $B$ is bounded on $L^2(\mathbb{R}^n)$ and then replace $v$ by $J^s u$.[/guided]

[step:Use symbolic composition to identify the conjugated operator as order zero] The Bessel symbols $j_{s-m}$ and $j_{-s}$ belong to $S^{s-m}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)$ and $S^{-s}_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)$, respectively, because they are independent of $x$ and their $\xi$-derivatives satisfy the defining symbol estimates. By the type $(1,0)$ pseudodifferential composition theorem, applied first to $J^{s-m}A$ and then to $(J^{s-m}A)J^{-s}$, the operator $B$ is a Kohn-Nirenberg pseudodifferential operator \begin{align*} B=\operatorname{Op}(b) \end{align*} with symbol $b\in S^0_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)$. We use the quantitative form of the composition theorem here. Its hypotheses are satisfied because $j_{s-m}\in S^{s-m}_{1,0}$, $a\in S^m_{1,0}$, and $j_{-s}\in S^{-s}_{1,0}$, so the composed orders add to $(s-m)+m+(-s)=0$. Hence, for every prescribed nonnegative integer $N_0$, there exist an integer $N_1=N_1(n,m,s,N_0)$ and a constant $C_1=C_1(n,m,s,N_0)>0$ such that \begin{align*} \sum_{|\alpha|+|\beta|\leq N_0}p_{\alpha,\beta}^{(0)}(b)\leq C_1\sum_{|\alpha|+|\beta|\leq N_1}p_{\alpha,\beta}^{(m)}(a). \end{align*} This is the quantitative composition theorem for type $(1,0)$ Kohn-Nirenberg pseudodifferential operators. [/step]

[step:Apply Calderón-Vaillancourt to the order-zero operator]By the [Calderón-Vaillancourt theorem](/theorems/7693) for order-zero Kohn-Nirenberg symbols on $\mathbb{R}^n$, there exist an integer $N_0=N_0(n)\in\mathbb{N}$ and a constant $C_0=C_0(n)>0$ such that every operator $\operatorname{Op}(b)$ with $b\in S^0_{1,0}$ satisfies \begin{align*} \|\operatorname{Op}(b)w\|_{L^2(\mathbb{R}^n)}\leq C_0\sum_{|\alpha|+|\beta|\leq N_0}p_{\alpha,\beta}^{(0)}(b)\|w\|_{L^2(\mathbb{R}^n)} \end{align*} for all $w\in\mathcal{S}(\mathbb{R}^n)$. Applying this to $B=\operatorname{Op}(b)$ and using the finite-seminorm bound from the composition step gives \begin{align*} \|Bw\|_{L^2(\mathbb{R}^n)}\leq C_0C_1\sum_{|\alpha|+|\beta|\leq N_1}p_{\alpha,\beta}^{(m)}(a)\|w\|_{L^2(\mathbb{R}^n)} \end{align*} for every $w\in\mathcal{S}(\mathbb{R}^n)$.[/step]

[guided]At this point the operator $B$ has order zero. More precisely, the composition step produced a symbol $b\in S^0_{1,0}(\mathbb{R}^n\times\mathbb{R}^n)$ and identified $B$ with the Kohn-Nirenberg operator $\operatorname{Op}(b)$ on $\mathcal{S}(\mathbb{R}^n)$. These are exactly the hypotheses needed for the Calderón-Vaillancourt theorem. The theorem says that an order-zero Kohn-Nirenberg operator is bounded on $L^2(\mathbb{R}^n)$, and more precisely that its $L^2$ operator norm is controlled by finitely many order-zero symbol seminorms. Thus there are an integer $N_0=N_0(n)$ and a constant $C_0=C_0(n)>0$ such that \begin{align*} \|\operatorname{Op}(b)w\|_{L^2(\mathbb{R}^n)}\leq C_0\sum_{|\alpha|+|\beta|\leq N_0}p_{\alpha,\beta}^{(0)}(b)\|w\|_{L^2(\mathbb{R}^n)} \end{align*} for all $w\in\mathcal{S}(\mathbb{R}^n)$. The remaining bookkeeping is important because the theorem statement asks for control by seminorms of the original symbol $a$, not by seminorms of the composed symbol $b$. The quantitative composition theorem already gives such a comparison: for the particular integer $N_0$ required by Calderón-Vaillancourt, there are $N_1=N_1(n,m,s)$ and $C_1=C_1(n,m,s)>0$ such that \begin{align*} \sum_{|\alpha|+|\beta|\leq N_0}p_{\alpha,\beta}^{(0)}(b)\leq C_1\sum_{|\alpha|+|\beta|\leq N_1}p_{\alpha,\beta}^{(m)}(a). \end{align*} Combining the two estimates gives \begin{align*} \|Bw\|_{L^2(\mathbb{R}^n)}\leq C_0C_1\sum_{|\alpha|+|\beta|\leq N_1}p_{\alpha,\beta}^{(m)}(a)\|w\|_{L^2(\mathbb{R}^n)}. \end{align*} This is the desired $L^2$ estimate for the conjugated operator.[/guided]

[step:Transfer the $L^2$ estimate back to the Sobolev estimate] Set \begin{align*} C_s(a):=C_0C_1\sum_{|\alpha|+|\beta|\leq N_1}p_{\alpha,\beta}^{(m)}(a). \end{align*} For $u\in\mathcal{S}(\mathbb{R}^n)$ and $v=J^s u$, the preceding step gives \begin{align*} \|Au\|_{H^{s-m}(\mathbb{R}^n)}=\|Bv\|_{L^2(\mathbb{R}^n)}\leq C_s(a)\|v\|_{L^2(\mathbb{R}^n)}. \end{align*} Since $v=J^s u$, we have \begin{align*} \|v\|_{L^2(\mathbb{R}^n)}=\|J^s u\|_{L^2(\mathbb{R}^n)}=\|u\|_{H^s(\mathbb{R}^n)}. \end{align*} Hence \begin{align*} \|Au\|_{H^{s-m}(\mathbb{R}^n)}\leq C_s(a)\|u\|_{H^s(\mathbb{R}^n)} \end{align*} for every $u\in\mathcal{S}(\mathbb{R}^n)$. [/step]

[step:Extend from Schwartz functions by density and prove uniqueness] By the density of Schwartz functions in Sobolev spaces, $\mathcal{S}(\mathbb{R}^n)$ is dense in $H^s(\mathbb{R}^n)$. Let $u\in H^s(\mathbb{R}^n)$, and choose a sequence $(u_k)_{k=1}^{\infty}$ in $\mathcal{S}(\mathbb{R}^n)$ such that \begin{align*} \|u_k-u\|_{H^s(\mathbb{R}^n)}\to 0. \end{align*} The estimate from the previous step implies \begin{align*} \|Au_k-Au_\ell\|_{H^{s-m}(\mathbb{R}^n)}\leq C_s(a)\|u_k-u_\ell\|_{H^s(\mathbb{R}^n)} \end{align*} for all $k,\ell\in\mathbb{N}$. Therefore $(Au_k)_{k=1}^{\infty}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in the [Hilbert space](/page/Hilbert%20Space) $H^{s-m}(\mathbb{R}^n)$, and hence it converges to some element of $H^{s-m}(\mathbb{R}^n)$. Define the extension $\widetilde{A}:H^s(\mathbb{R}^n)\to H^{s-m}(\mathbb{R}^n)$ by \begin{align*} \widetilde{A}u:=\lim_{k\to\infty}Au_k. \end{align*} This definition is independent of the approximating sequence. Indeed, if $(w_k)_{k=1}^{\infty}$ is another sequence in $\mathcal{S}(\mathbb{R}^n)$ converging to $u$ in $H^s(\mathbb{R}^n)$, then \begin{align*} \|Au_k-Aw_k\|_{H^{s-m}(\mathbb{R}^n)}\leq C_s(a)\|u_k-w_k\|_{H^s(\mathbb{R}^n)}. \end{align*} The right-hand side tends to $0$, so the two limits agree. The same estimate passes to the limit and gives \begin{align*} \|\widetilde{A}u\|_{H^{s-m}(\mathbb{R}^n)}\leq C_s(a)\|u\|_{H^s(\mathbb{R}^n)}. \end{align*} Thus $\widetilde{A}$ is a bounded linear extension of $A$. Linearity follows from the definition by limits and the linearity of $A$ on $\mathcal{S}(\mathbb{R}^n)$: if $u,w\in H^s(\mathbb{R}^n)$, $\lambda,\mu\in\mathbb{C}$, and $u_k,w_k\in\mathcal{S}(\mathbb{R}^n)$ converge to $u,w$ in $H^s(\mathbb{R}^n)$, then $\lambda u_k+\mu w_k\to \lambda u+\mu w$ in $H^s(\mathbb{R}^n)$ and \begin{align*} \widetilde{A}(\lambda u+\mu w)=\lim_{k\to\infty}A(\lambda u_k+\mu w_k)=\lambda\widetilde{A}u+\mu\widetilde{A}w. \end{align*} Finally, suppose $T:H^s(\mathbb{R}^n)\to H^{s-m}(\mathbb{R}^n)$ is another [bounded linear operator](/page/Bounded%20Linear%20Operator) agreeing with $A$ on $\mathcal{S}(\mathbb{R}^n)$. For $u\in H^s(\mathbb{R}^n)$ and $u_k\to u$ in $H^s(\mathbb{R}^n)$ with $u_k\in\mathcal{S}(\mathbb{R}^n)$, continuity gives \begin{align*} Tu=\lim_{k\to\infty}Tu_k=\lim_{k\to\infty}Au_k=\widetilde{A}u. \end{align*} Therefore $T=\widetilde{A}$, proving uniqueness. With $N:=N_1$ and $C:=C_0C_1$, the finite-seminorm operator norm bound in the statement follows. [/step]