[proofplan] We prove continuity by estimating every Schwartz seminorm of $\operatorname{Op}(a)u$. Differentiating the oscillatory integral in $x$ produces finitely many terms in which derivatives fall either on the symbol or on the phase, the latter contributing polynomial powers of $\xi$. Multiplication by $x^\beta$ is converted into $\xi$-derivatives of the phase, and [integration by parts](/theorems/210) in $\xi$ transfers these derivatives onto the product of the symbol factor and $\hat{u}$. Symbol estimates and rapid decay of $\hat{u}$ then give a uniform bound by finitely many Schwartz seminorms of $u$. [/proofplan]

[step:Fix the seminorms and symbol bounds to be estimated] Let $\mathbb{N}_0 := \{0,1,2,\dots\}$, so that $\mathbb{N}_0^n$ denotes the set of $n$-dimensional multi-indices with non-negative integer components. For multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, define the Schwartz seminorm \begin{align*} p_{\alpha,\beta}: \mathcal S(\mathbb R^n)\to[0,\infty), \qquad p_{\alpha,\beta}(v) := \sup_{x \in \mathbb{R}^n} |x^\beta D^\alpha v(x)| \end{align*} for every $v \in \mathcal{S}(\mathbb{R}^n)$, where $x^\beta := x_1^{\beta_1}\cdots x_n^{\beta_n}$ and $D^\alpha := \partial_{x_1}^{\alpha_1}\cdots \partial_{x_n}^{\alpha_n}$. Since $a \in S^m_{1,0}(\mathbb{R}^n_x \times \mathbb{R}^n_\xi)$, for every pair of multi-indices $\gamma,\rho \in \mathbb{N}_0^n$ there is a finite constant \begin{align*} C_{\gamma,\rho}(a) := \sup_{(x,\xi)\in \mathbb{R}^n\times \mathbb{R}^n} \langle \xi\rangle^{-m+|\rho|} |\partial_x^\gamma \partial_\xi^\rho a(x,\xi)| \end{align*} where $\langle \xi\rangle := (1+|\xi|^2)^{1/2}$. Thus \begin{align*} |\partial_x^\gamma \partial_\xi^\rho a(x,\xi)| \le C_{\gamma,\rho}(a)\langle \xi\rangle^{m-|\rho|} \end{align*} for all $x,\xi \in \mathbb{R}^n$. For $N \in \mathbb{N}_0$ and a smooth function $w: \mathbb{R}^n \to \mathbb{C}$, define the Fourier-side Schwartz seminorm \begin{align*} q_{N,\tau}: C^\infty(\mathbb R^n)\to[0,\infty], \qquad q_{N,\tau}(w) := \sup_{\xi \in \mathbb{R}^n}\langle \xi\rangle^N |\partial_\xi^\tau w(\xi)| \end{align*} for each multi-index $\tau \in \mathbb{N}_0^n$. [/step]

[step:Differentiate the operator under the Fourier integral]Fix multi-indices $\alpha,\beta \in \mathbb{N}_0^n$. Use the Fourier-side seminorms $q_{N,\tau}:C^\infty(\mathbb R^n)\to[0,\infty]$ defined in the previous step. For each decomposition $\gamma+\delta=\alpha$, the symbol estimate gives \begin{align*} |\xi^\delta\partial_x^\gamma a(x,\xi)\hat{u}(\xi)| \le C_{\gamma,0}(a)q_{N,0}(\hat{u})\langle \xi\rangle^{m+|\delta|-N} \end{align*} for every integer $N \in \mathbb{N}_0$. Choose $N>n+\max\{0,m+|\alpha|\}$. Since $|\delta|\le |\alpha|$, the right-hand side is bounded by the integrable function $C_{\gamma,0}(a)q_{N,0}(\hat{u})\langle \xi\rangle^{m+|\alpha|-N}$ on $\mathbb{R}^n$. The same estimate, with $\gamma$ enlarged by the additional $x$-derivatives being taken, dominates each differentiated integrand. Therefore differentiation under the integral is justified by the [dominated convergence theorem](/theorems/4). Leibniz' rule gives \begin{align*} D^\alpha_x\left(e^{i x\cdot \xi}a(x,\xi)\right)=\sum_{\gamma+\delta=\alpha} c_{\gamma,\delta}\, e^{i x\cdot \xi}\xi^\delta\,\partial_x^\gamma a(x,\xi) \end{align*} for constants $c_{\gamma,\delta}\in \mathbb{C}$ depending only on $\gamma$ and $\delta$. Hence \begin{align*} D^\alpha \operatorname{Op}(a)u(x)=(2\pi)^{-n/2}\sum_{\gamma+\delta=\alpha} c_{\gamma,\delta}\int_{\mathbb{R}^n} e^{i x\cdot \xi}\xi^\delta \partial_x^\gamma a(x,\xi)\hat{u}(\xi)\,d\mathcal{L}^n(\xi). \end{align*}[/step]

[guided]We first isolate what happens when an $x$-derivative hits the oscillatory integral. The derivative $\partial_{x_j}$ can fall either on the exponential factor $e^{ix\cdot \xi}$ or on the symbol $a(x,\xi)$. If it hits the exponential, then \begin{align*} \partial_{x_j}e^{ix\cdot \xi}=i\xi_j e^{ix\cdot \xi}. \end{align*} Thus each derivative landing on the phase produces one factor of $\xi_j$, while each derivative landing on the symbol produces one $x$-derivative of $a$. Applying the multi-index Leibniz rule to $D_x^\alpha(e^{ix\cdot \xi}a(x,\xi))$ gives \begin{align*} D^\alpha_x\left(e^{i x\cdot \xi}a(x,\xi)\right)=\sum_{\gamma+\delta=\alpha} c_{\gamma,\delta}\, e^{i x\cdot \xi}\xi^\delta\,\partial_x^\gamma a(x,\xi). \end{align*} Here $\gamma$ records the derivatives that fall on $a$, and $\delta$ records the derivatives that fall on the phase. The formula may be inserted under the integral because $u\in \mathcal{S}(\mathbb{R}^n)$ implies that $\hat{u}$ decays faster than every power of $|\xi|$, while each $\partial_x^\gamma a(x,\xi)$ grows at most like a fixed power of $\langle \xi\rangle$. More explicitly, choosing an integer $N>n+\max\{0,m+|\alpha|\}$ gives an integrable majorant proportional to $\langle \xi\rangle^{m+|\alpha|-N}$ for every term arising from $D_x^\alpha$. Therefore every differentiated integrand is absolutely integrable in $\xi$, and the [dominated convergence theorem](/theorems/7529) applies. We obtain \begin{align*} D^\alpha \operatorname{Op}(a)u(x)=(2\pi)^{-n/2}\sum_{\gamma+\delta=\alpha} c_{\gamma,\delta}\int_{\mathbb{R}^n} e^{i x\cdot \xi}\xi^\delta \partial_x^\gamma a(x,\xi)\hat{u}(\xi)\,d\mathcal{L}^n(\xi). \end{align*}[/guided]

[step:Trade the polynomial weight in $x$ for derivatives in $\xi$] For a fixed pair $\gamma,\delta$ with $\gamma+\delta=\alpha$, define \begin{align*} b_{\gamma,\delta}: \mathbb{R}^n_x \times \mathbb{R}^n_\xi \to \mathbb{C}, \qquad b_{\gamma,\delta}(x,\xi):=\xi^\delta \partial_x^\gamma a(x,\xi). \end{align*} The identity \begin{align*} x^\beta e^{ix\cdot \xi}=(-i)^{|\beta|}\partial_\xi^\beta e^{ix\cdot \xi} \end{align*} allows [integration by parts](/theorems/2098) in $\xi$. Since $\hat{u}$ and all its derivatives are rapidly decreasing and $b_{\gamma,\delta}$ has polynomial growth in $\xi$, the boundary terms vanish. Thus \begin{align*} x^\beta\int_{\mathbb{R}^n} e^{ix\cdot \xi}b_{\gamma,\delta}(x,\xi)\hat{u}(\xi)\,d\mathcal{L}^n(\xi)=i^{|\beta|}\int_{\mathbb{R}^n} e^{ix\cdot \xi}\partial_\xi^\beta\left(b_{\gamma,\delta}(x,\xi)\hat{u}(\xi)\right)\,d\mathcal{L}^n(\xi). \end{align*} Expanding the $\xi$-derivative by Leibniz' rule gives \begin{align*} \partial_\xi^\beta\left(b_{\gamma,\delta}(x,\xi)\hat{u}(\xi)\right)=\sum_{\rho+\tau=\beta} d_{\rho,\tau}\,\partial_\xi^\rho b_{\gamma,\delta}(x,\xi)\partial_\xi^\tau\hat{u}(\xi) \end{align*} for constants $d_{\rho,\tau}\in\mathbb{C}$ depending only on $\rho$ and $\tau$. [/step]

[step:Estimate every term by finitely many Schwartz seminorms of $u$]Fix $\rho,\tau \in \mathbb{N}_0^n$ with $\rho+\tau=\beta$. By Leibniz' rule applied to $b_{\gamma,\delta}(x,\xi)=\xi^\delta\partial_x^\gamma a(x,\xi)$, there are constants $e_{\kappa,\lambda}\in\mathbb{C}$ such that \begin{align*} \partial_\xi^\rho b_{\gamma,\delta}(x,\xi)=\sum_{\kappa+\lambda=\rho} e_{\kappa,\lambda}\,\partial_\xi^\kappa(\xi^\delta)\,\partial_x^\gamma\partial_\xi^\lambda a(x,\xi). \end{align*} If $\partial_\xi^\kappa(\xi^\delta)$ is nonzero, then \begin{align*} |\partial_\xi^\kappa(\xi^\delta)| \le A_{\delta,\kappa}\langle \xi\rangle^{|\delta|-|\kappa|} \end{align*} for a constant $A_{\delta,\kappa}\ge 0$. The symbol estimate gives \begin{align*} |\partial_x^\gamma\partial_\xi^\lambda a(x,\xi)|\le C_{\gamma,\lambda}(a)\langle \xi\rangle^{m-|\lambda|}. \end{align*} Since $\kappa+\lambda=\rho$, the product satisfies \begin{align*} |\partial_\xi^\rho b_{\gamma,\delta}(x,\xi)|\le B_{\gamma,\delta,\rho}(a)\langle \xi\rangle^{m+|\delta|-|\rho|} \end{align*} for a finite constant $B_{\gamma,\delta,\rho}(a)$ depending only on finitely many symbol seminorms of $a$. Choose an integer $N_{\alpha,m}>n+\max\{0,m+|\alpha|\}$. Since $|\delta|\le |\alpha|$, the preceding estimate yields \begin{align*} |\partial_\xi^\rho b_{\gamma,\delta}(x,\xi)\partial_\xi^\tau\hat{u}(\xi)|\le B_{\gamma,\delta,\rho}(a)q_{N_{\alpha,m},\tau}(\hat{u})\langle \xi\rangle^{m+|\alpha|-N_{\alpha,m}}. \end{align*} The exponent satisfies $m+|\alpha|-N_{\alpha,m}<-n$, so \begin{align*} I_{\alpha,m}:=\int_{\mathbb{R}^n}\langle \xi\rangle^{m+|\alpha|-N_{\alpha,m}}\,d\mathcal{L}^n(\xi)<\infty. \end{align*} Combining the finitely many terms from the preceding steps gives \begin{align*} p_{\alpha,\beta}(\operatorname{Op}(a)u)\le K_{\alpha,\beta,m}(a)\sum_{|\tau|\le |\beta|}q_{N_{\alpha,m},\tau}(\hat{u}) \end{align*} where $K_{\alpha,\beta,m}(a)<\infty$ depends only on $\alpha,\beta,m,n$ and finitely many symbol seminorms $C_{\gamma,\lambda}(a)$.[/step]

[guided]The goal is to obtain a bound uniform in $x$. After differentiating in $x$ and integrating by parts in $\xi$, each term has the form \begin{align*} \int_{\mathbb{R}^n} e^{ix\cdot \xi}\partial_\xi^\rho b_{\gamma,\delta}(x,\xi)\partial_\xi^\tau\hat{u}(\xi)\,d\mathcal{L}^n(\xi). \end{align*} The exponential has absolute value $1$, so it remains to prove that the rest is integrable in $\xi$ with a bound independent of $x$. We estimate $\partial_\xi^\rho b_{\gamma,\delta}$. Since \begin{align*} b_{\gamma,\delta}(x,\xi)=\xi^\delta\partial_x^\gamma a(x,\xi), \end{align*} Leibniz' rule gives \begin{align*} \partial_\xi^\rho b_{\gamma,\delta}(x,\xi)=\sum_{\kappa+\lambda=\rho} e_{\kappa,\lambda}\,\partial_\xi^\kappa(\xi^\delta)\,\partial_x^\gamma\partial_\xi^\lambda a(x,\xi). \end{align*} The derivative of the monomial $\xi^\delta$ is bounded by a polynomial: \begin{align*} |\partial_\xi^\kappa(\xi^\delta)| \le A_{\delta,\kappa}\langle \xi\rangle^{|\delta|-|\kappa|}. \end{align*} The symbol estimate for $a\in S^m_{1,0}$ gives \begin{align*} |\partial_x^\gamma\partial_\xi^\lambda a(x,\xi)|\le C_{\gamma,\lambda}(a)\langle \xi\rangle^{m-|\lambda|}. \end{align*} Multiplying these bounds and using $\kappa+\lambda=\rho$ gives the uniform estimate \begin{align*} |\partial_\xi^\rho b_{\gamma,\delta}(x,\xi)|\le B_{\gamma,\delta,\rho}(a)\langle \xi\rangle^{m+|\delta|-|\rho|}. \end{align*} Now we use the rapid decay of $\hat{u}$. Choose an integer \begin{align*} N_{\alpha,m}>n+\max\{0,m+|\alpha|\}. \end{align*} Because $|\delta|\le |\alpha|$, the Fourier-side seminorm $q_{N_{\alpha,m},\tau}(\hat{u})$ gives \begin{align*} |\partial_\xi^\tau\hat{u}(\xi)|\le q_{N_{\alpha,m},\tau}(\hat{u})\langle \xi\rangle^{-N_{\alpha,m}}. \end{align*} Therefore \begin{align*} |\partial_\xi^\rho b_{\gamma,\delta}(x,\xi)\partial_\xi^\tau\hat{u}(\xi)|\le B_{\gamma,\delta,\rho}(a)q_{N_{\alpha,m},\tau}(\hat{u})\langle \xi\rangle^{m+|\alpha|-N_{\alpha,m}}. \end{align*} The choice of $N_{\alpha,m}$ makes the exponent strictly smaller than $-n$, so the function $\xi\mapsto \langle \xi\rangle^{m+|\alpha|-N_{\alpha,m}}$ is integrable over $\mathbb{R}^n$ with respect to $\mathcal{L}^n$. Hence \begin{align*} I_{\alpha,m}:=\int_{\mathbb{R}^n}\langle \xi\rangle^{m+|\alpha|-N_{\alpha,m}}\,d\mathcal{L}^n(\xi)<\infty. \end{align*} Summing over the finitely many multi-indices produced by Leibniz' rule and the integration by parts identity gives \begin{align*} p_{\alpha,\beta}(\operatorname{Op}(a)u)\le K_{\alpha,\beta,m}(a)\sum_{|\tau|\le |\beta|}q_{N_{\alpha,m},\tau}(\hat{u}). \end{align*} This is the essential estimate: every output Schwartz seminorm is controlled by finitely many seminorms of $\hat{u}$ and finitely many symbol seminorms of $a$.[/guided]

[step:Convert Fourier-side seminorms into Schwartz seminorms of $u$]For each pair $N,\tau$, the seminorm $q_{N,\tau}(\hat{u})$ is controlled by finitely many Schwartz seminorms of $u$. Choose a finite constant $A_N>0$ such that \begin{align*} \langle \xi\rangle^N \le A_N\sum_{|\eta|\le N+n+1}|\xi^\eta| \end{align*} for all $\xi\in\mathbb{R}^n$. For every multi-index $\eta$, differentiation under the defining Fourier integral gives \begin{align*} \partial_\xi^\tau\hat{u}(\xi)=c_\tau\int_{\mathbb{R}^n}e^{-iy\cdot \xi}y^\tau u(y)\,d\mathcal{L}^n(y) \end{align*} for a constant $c_\tau\in\mathbb{C}$. Multiplication by $\xi^\eta$ is obtained by integrating by parts in the $y$ variables, so \begin{align*} \xi^\eta\partial_\xi^\tau\hat{u}(\xi)=c_{\eta,\tau}\int_{\mathbb{R}^n}e^{-iy\cdot \xi}D_y^\eta(y^\tau u(y))\,d\mathcal{L}^n(y) \end{align*} for a constant $c_{\eta,\tau}\in\mathbb{C}$. Hence \begin{align*} |\xi^\eta\partial_\xi^\tau\hat{u}(\xi)|\le |c_{\eta,\tau}|\int_{\mathbb{R}^n}|D_y^\eta(y^\tau u(y))|\,d\mathcal{L}^n(y). \end{align*} Expanding $D_y^\eta(y^\tau u(y))$ by Leibniz' rule gives a finite sum of terms of the form $P(y)D_y^\mu u(y)$, where $P$ is a polynomial depending only on $\eta$ and $\tau$. Let $d=\deg P$ and choose an integer $R>d+n$. Since $|P(y)|\le C_P\langle y\rangle^d$, we have \begin{align*} \int_{\mathbb{R}^n}|P(y)D_y^\mu u(y)|\,d\mathcal{L}^n(y)\le C_P\left(\sup_{z\in\mathbb{R}^n}\langle z\rangle^R|D_z^\mu u(z)|\right)\int_{\mathbb{R}^n}\langle y\rangle^{d-R}\,d\mathcal{L}^n(y). \end{align*} The integral is finite because $d-R<-n$. Also, for a finite set $E_R\subset\mathbb{N}_0^n$ and a constant $C_R>0$, \begin{align*} \langle z\rangle^R\le C_R\sum_{\nu\in E_R}|z^\nu| \end{align*} for all $z\in\mathbb{R}^n$, after including the constant monomial in $E_R$. Hence \begin{align*} \sup_{z\in\mathbb{R}^n}\langle z\rangle^R|D_z^\mu u(z)|\le C_R\sum_{\nu\in E_R}p_{\mu,\nu}(u). \end{align*} Thus there exist finitely many multi-indices $\mu,\nu$ and a constant $L_{N,\tau}>0$ such that \begin{align*} q_{N,\tau}(\hat{u})\le L_{N,\tau}\sum_{\mu,\nu}p_{\mu,\nu}(u). \end{align*} Substituting this estimate into the previous bound gives, for every $\alpha,\beta$, \begin{align*} p_{\alpha,\beta}(\operatorname{Op}(a)u)\le M_{\alpha,\beta,m}(a)\sum_{\mu,\nu\in F_{\alpha,\beta}}p_{\mu,\nu}(u) \end{align*} where $F_{\alpha,\beta}$ is a finite set of pairs of multi-indices and $M_{\alpha,\beta,m}(a)<\infty$ depends on finitely many symbol seminorms of $a$.[/step]

[guided]We now justify in detail why the seminorms of $\hat{u}$ that appeared in the estimate are controlled by ordinary Schwartz seminorms of $u$. The factor $\langle \xi\rangle^N$ is not itself a monomial, but it is bounded above by finitely many polynomial weights: choose a finite constant $A_N>0$ such that \begin{align*} \langle \xi\rangle^N \le A_N\sum_{|\eta|\le N+n+1}|\xi^\eta| \end{align*} for all $\xi\in\mathbb{R}^n$. Therefore it is enough to control $\sup_\xi |\xi^\eta\partial_\xi^\tau\hat{u}(\xi)|$ for finitely many $\eta$. By differentiating the [Fourier transform](/page/Fourier%20Transform) under the integral, which is justified because $y^\tau u(y)$ is integrable with respect to $\mathcal{L}^n$, we have \begin{align*} \partial_\xi^\tau\hat{u}(\xi)=c_\tau\int_{\mathbb{R}^n}e^{-iy\cdot \xi}y^\tau u(y)\,d\mathcal{L}^n(y) \end{align*} for a constant $c_\tau\in\mathbb{C}$. To handle the additional factor $\xi^\eta$, integrate by parts in the $y$ variables. The boundary terms vanish because every derivative of $y^\tau u(y)$ is rapidly decreasing. This gives \begin{align*} \xi^\eta\partial_\xi^\tau\hat{u}(\xi)=c_{\eta,\tau}\int_{\mathbb{R}^n}e^{-iy\cdot \xi}D_y^\eta(y^\tau u(y))\,d\mathcal{L}^n(y) \end{align*} for a constant $c_{\eta,\tau}\in\mathbb{C}$. Taking absolute values removes the oscillatory factor, since $|e^{-iy\cdot\xi}|=1$, and yields \begin{align*} |\xi^\eta\partial_\xi^\tau\hat{u}(\xi)|\le |c_{\eta,\tau}|\int_{\mathbb{R}^n}|D_y^\eta(y^\tau u(y))|\,d\mathcal{L}^n(y). \end{align*} It remains to bound this $L^1$ norm by Schwartz seminorms of $u$. Leibniz' rule expands $D_y^\eta(y^\tau u(y))$ into finitely many terms $P(y)D_y^\mu u(y)$, with $P$ a polynomial determined by $\eta$ and $\tau$. Let $d=\deg P$ and choose an integer $R>d+n$. Since $|P(y)|\le C_P\langle y\rangle^d$, \begin{align*} \int_{\mathbb{R}^n}|P(y)D_y^\mu u(y)|\,d\mathcal{L}^n(y)\le C_P\left(\sup_{z\in\mathbb{R}^n}\langle z\rangle^R|D_z^\mu u(z)|\right)\int_{\mathbb{R}^n}\langle y\rangle^{d-R}\,d\mathcal{L}^n(y). \end{align*} The last integral is finite. The weighted supremum is controlled by finitely many Schwartz seminorms: choose a finite set $E_R\subset\mathbb{N}_0^n$ containing the zero multi-index and a constant $C_R>0$ such that \begin{align*} \langle z\rangle^R\le C_R\sum_{\nu\in E_R}|z^\nu|. \end{align*} Then \begin{align*} \sup_{z\in\mathbb{R}^n}\langle z\rangle^R|D_z^\mu u(z)|\le C_R\sum_{\nu\in E_R}p_{\mu,\nu}(u). \end{align*} Summing the finitely many Leibniz terms gives \begin{align*} q_{N,\tau}(\hat{u})\le L_{N,\tau}\sum_{\mu,\nu}p_{\mu,\nu}(u). \end{align*} Substituting this into the previous estimate gives \begin{align*} p_{\alpha,\beta}(\operatorname{Op}(a)u)\le M_{\alpha,\beta,m}(a)\sum_{\mu,\nu\in F_{\alpha,\beta}}p_{\mu,\nu}(u). \end{align*} This proves that the Fourier-side seminorms required by the symbol estimate cost only finitely many Schwartz seminorms of the original input $u$.[/guided]

[step:Conclude Schwartz regularity and continuity]The preceding estimate shows that every Schwartz seminorm of $\operatorname{Op}(a)u$ is finite whenever $u\in\mathcal{S}(\mathbb{R}^n)$. Therefore $\operatorname{Op}(a)u\in\mathcal{S}(\mathbb{R}^n)$. The same estimate also proves continuity. For each seminorm $p_{\alpha,\beta}$ on the target space, there are finitely many source seminorms $p_{\mu,\nu}$ and a constant $M_{\alpha,\beta,m}(a)$ such that \begin{align*} p_{\alpha,\beta}(\operatorname{Op}(a)u)\le M_{\alpha,\beta,m}(a)\sum_{\mu,\nu\in F_{\alpha,\beta}}p_{\mu,\nu}(u) \end{align*} for all $u\in\mathcal{S}(\mathbb{R}^n)$. This is exactly continuity for a [linear map](/page/Linear%20Map) between locally convex spaces whose topologies are generated by these seminorms. Hence \begin{align*} \operatorname{Op}(a):\mathcal{S}(\mathbb{R}^n)\to\mathcal{S}(\mathbb{R}^n) \end{align*} is continuous.[/step]

[guided]The final estimate has the precise form required for continuity in the Schwartz topology. The topology on $\mathcal{S}(\mathbb{R}^n)$ is generated by the seminorms $p_{\alpha,\beta}$. Therefore, to prove that the linear map $\operatorname{Op}(a)$ is continuous, it is enough to show that each target seminorm of $\operatorname{Op}(a)u$ is bounded by a finite linear combination of source seminorms of $u$. For the fixed target seminorm $p_{\alpha,\beta}$, the preceding step gives a finite set $F_{\alpha,\beta}$ of pairs of multi-indices and a finite constant $M_{\alpha,\beta,m}(a)$ such that \begin{align*} p_{\alpha,\beta}(\operatorname{Op}(a)u)\le M_{\alpha,\beta,m}(a)\sum_{\mu,\nu\in F_{\alpha,\beta}}p_{\mu,\nu}(u) \end{align*} for every $u\in\mathcal{S}(\mathbb{R}^n)$. This finite-seminorm domination proves continuity of $\operatorname{Op}(a)$ from $\mathcal{S}(\mathbb{R}^n)$ to itself. It also shows that each $p_{\alpha,\beta}(\operatorname{Op}(a)u)$ is finite, so $\operatorname{Op}(a)u$ is a Schwartz function.[/guided]