[proofplan] We reduce the Sobolev estimate to an $L^2$ estimate by conjugating $A$ with Bessel potential operators. The symbolic composition theorem shows that the conjugated operator $J^{s-m}AJ^{-s}$ has an order-zero symbol with global uniform estimates. The global [Calderón-Vaillancourt theorem](/theorems/7693) then gives $L^2$ boundedness, which is exactly the desired $H^s \to H^{s-m}$ bound after undoing the Bessel potentials. Finally, density of $\mathcal{S}(\mathbb{R}^n)$ in $H^s(\mathbb{R}^n)$ gives the unique continuous extension. [/proofplan]

[step:Introduce Bessel potentials and identify Sobolev norms with $L^2$ norms] Let $\mathcal{F}: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ denote the [Fourier transform](/page/Fourier%20Transform) with symmetric normalization, and let $\mathcal{F}^{-1}: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ denote its inverse. For $u \in \mathcal{S}(\mathbb{R}^n)$, write $\widehat{u}:=\mathcal{F}u$. For each $t \in \mathbb{R}$, define the Bessel potential operator $J^t: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ by \begin{align*} J^t u := \mathcal{F}^{-1}(\langle \xi\rangle^t \widehat{u}(\xi)). \end{align*} Equivalently, $J^t = \operatorname{Op}(j_t)$, where the symbol $j_t: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ is the map defined by \begin{align*} j_t(x,\xi) := \langle \xi\rangle^t. \end{align*} The symbol $j_t$ is independent of $x$ and belongs to $S^t_{1,0}(\mathbb{R}^n \times \mathbb{R}^n)$, because each $\xi$-derivative lowers the order by the corresponding derivative order. For $r \in \mathbb{R}$, the Sobolev norm on $H^r(\mathbb{R}^n)$ is \begin{align*} \|u\|_{H^r(\mathbb{R}^n)} := \|J^r u\|_{L^2(\mathbb{R}^n)}. \end{align*} Thus $J^r: H^r(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ is an isometric isomorphism, with inverse $J^{-r}: L^2(\mathbb{R}^n) \to H^r(\mathbb{R}^n)$. This uses the standard Bessel-potential isomorphism on Sobolev spaces, cited here as a result not yet in the wiki: Bessel potential isomorphism on Sobolev spaces. [/step]

[step:Conjugate the operator to obtain an order-zero pseudodifferential operator]Fix $s \in \mathbb{R}$. Define the conjugated operator $B: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ by \begin{align*} B := J^{s-m} A J^{-s}. \end{align*} Since $J^{s-m} = \operatorname{Op}(j_{s-m})$, $A = \operatorname{Op}(a)$, and $J^{-s} = \operatorname{Op}(j_{-s})$, the global pseudodifferential composition theorem applies to the symbols $j_{s-m} \in S^{s-m}_{1,0}$, $a \in S^m_{1,0}$, and $j_{-s} \in S^{-s}_{1,0}$. This theorem is cited here as a result not yet in the wiki: Composition theorem for pseudodifferential operators. The composition theorem gives a symbol $b: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ such that \begin{align*} B = \operatorname{Op}(b) \end{align*} on $\mathcal{S}(\mathbb{R}^n)$, and the order of $b$ is \begin{align*} (s-m) + m + (-s) = 0. \end{align*} Therefore $b \in S^0_{1,0}(\mathbb{R}^n \times \mathbb{R}^n)$. More explicitly, for every pair of multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, there exists a constant $C'_{\alpha,\beta,s} > 0$ such that \begin{align*} |\partial_x^\beta \partial_\xi^\alpha b(x,\xi)| \leq C'_{\alpha,\beta,s}\langle \xi\rangle^{-|\alpha|} \end{align*} for all $(x,\xi) \in \mathbb{R}^n \times \mathbb{R}^n$.[/step]

[guided]The purpose of the conjugation is to turn the desired Sobolev estimate into an $L^2$ estimate. We define the conjugated operator $B: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ by \begin{align*} B := J^{s-m} A J^{-s}. \end{align*} If $B$ is bounded on $L^2(\mathbb{R}^n)$, then applying $B$ to $J^s u$ will control $J^{s-m}Au$, which is precisely the $H^{s-m}$ norm of $Au$. We now check that $B$ is an order-zero pseudodifferential operator. The operator $J^{s-m}$ has symbol $j_{s-m}(x,\xi) = \langle \xi\rangle^{s-m}$, so $j_{s-m} \in S^{s-m}_{1,0}$. The operator $A$ has symbol $a \in S^m_{1,0}$ by hypothesis. The operator $J^{-s}$ has symbol $j_{-s}(x,\xi) = \langle \xi\rangle^{-s}$, so $j_{-s} \in S^{-s}_{1,0}$. We apply the global composition theorem for Kohn-Nirenberg pseudodifferential operators, cited here as a result not yet in the wiki: Composition theorem for pseudodifferential operators. Its hypotheses require symbols in global $S^{r}_{1,0}$ classes with uniform estimates in $x$ and $\xi$. These hypotheses hold because $j_{s-m}$ and $j_{-s}$ are Bessel symbols with standard derivative bounds, and $a$ satisfies the stated global uniform estimates. The theorem gives a symbol $b: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ such that \begin{align*} B = \operatorname{Op}(b) \end{align*} on $\mathcal{S}(\mathbb{R}^n)$. The order of the composed symbol is the sum of the three orders: \begin{align*} (s-m) + m + (-s) = 0. \end{align*} Thus $b \in S^0_{1,0}(\mathbb{R}^n \times \mathbb{R}^n)$. In concrete terms, this means that for every pair of multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, there is a constant $C'_{\alpha,\beta,s} > 0$ such that \begin{align*} |\partial_x^\beta \partial_\xi^\alpha b(x,\xi)| \leq C'_{\alpha,\beta,s}\langle \xi\rangle^{-|\alpha|} \end{align*} for all $(x,\xi) \in \mathbb{R}^n \times \mathbb{R}^n$. This order-zero estimate is the exact input needed for Calderón-Vaillancourt.[/guided]

[step:Apply Calderón-Vaillancourt to the order-zero conjugated operator] The global Calderón-Vaillancourt theorem, cited here as a result not yet in the wiki: Calderón-Vaillancourt theorem, states that if $b \in S^0_{1,0}(\mathbb{R}^n \times \mathbb{R}^n)$ satisfies global uniform order-zero symbol estimates, then $\operatorname{Op}(b)$ extends to a bounded operator on $L^2(\mathbb{R}^n)$. The symbol $b$ obtained above satisfies precisely these estimates. Hence there exists a constant $C_{s} > 0$, depending on finitely many order-zero seminorms of $b$. Through the preceding composition theorem, these seminorms depend only on $s$, $m$, $n$, and finitely many of the symbol seminorms $C_{\alpha,\beta}$ of $a$. Therefore \begin{align*} \|Bv\|_{L^2(\mathbb{R}^n)} \leq C_s \|v\|_{L^2(\mathbb{R}^n)} \end{align*} for every $v \in \mathcal{S}(\mathbb{R}^n)$. [/step]

[step:Translate the $L^2$ bound back to the Sobolev estimate] Let $u \in \mathcal{S}(\mathbb{R}^n)$ and define \begin{align*} v := J^s u. \end{align*} Since $J^s$ maps $\mathcal{S}(\mathbb{R}^n)$ into itself, we have $v \in \mathcal{S}(\mathbb{R}^n)$. By the definition of $B$, \begin{align*} Bv = J^{s-m} A J^{-s} J^s u = J^{s-m}Au. \end{align*} Using the $L^2$ bound for $B$ and the definition of the Sobolev norm, \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*} The Calderón-Vaillancourt bound gives \begin{align*} \|Bv\|_{L^2(\mathbb{R}^n)} \leq C_s \|v\|_{L^2(\mathbb{R}^n)}. \end{align*} Since $v = J^s u$, \begin{align*} \|v\|_{L^2(\mathbb{R}^n)} = \|J^s u\|_{L^2(\mathbb{R}^n)} = \|u\|_{H^s(\mathbb{R}^n)}. \end{align*} Combining the preceding identities and inequality yields \begin{align*} \|Au\|_{H^{s-m}(\mathbb{R}^n)} \leq C_s \|u\|_{H^s(\mathbb{R}^n)}. \end{align*} [/step]

[step:Extend from Schwartz functions by density] The [Schwartz space](/page/Schwartz%20Space) $\mathcal{S}(\mathbb{R}^n)$ is dense in $H^s(\mathbb{R}^n)$, cited here as a result not yet in the wiki: Density of Schwartz functions in Sobolev spaces. 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 proved above gives, for all $k,\ell \in \mathbb{N}$, \begin{align*} \|Au_k - Au_\ell\|_{H^{s-m}(\mathbb{R}^n)} \leq C_s \|u_k - u_\ell\|_{H^s(\mathbb{R}^n)}. \end{align*} Thus $(Au_k)_{k=1}^\infty$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in $H^{s-m}(\mathbb{R}^n)$. Since $H^{s-m}(\mathbb{R}^n)$ is complete, there exists an element $w \in H^{s-m}(\mathbb{R}^n)$ such that \begin{align*} \|Au_k - w\|_{H^{s-m}(\mathbb{R}^n)} \to 0. \end{align*} Define the extension of $A$ by \begin{align*} Au := w. \end{align*} If $(\tilde{u}_k)_{k=1}^\infty$ is another Schwartz sequence converging to $u$ in $H^s(\mathbb{R}^n)$, the same estimate applied to $u_k - \tilde{u}_k$ gives \begin{align*} \|Au_k - A\tilde{u}_k\|_{H^{s-m}(\mathbb{R}^n)} \leq C_s \|u_k - \tilde{u}_k\|_{H^s(\mathbb{R}^n)} \to 0. \end{align*} Therefore the definition of $Au$ is independent of the approximating sequence. Passing to the limit in the Schwartz estimate gives \begin{align*} \|Au\|_{H^{s-m}(\mathbb{R}^n)} \leq C_s \|u\|_{H^s(\mathbb{R}^n)}. \end{align*} Hence $A$ extends uniquely to a [bounded linear operator](/page/Bounded%20Linear%20Operator) from $H^s(\mathbb{R}^n)$ to $H^{s-m}(\mathbb{R}^n)$, as claimed. [/step]