Proof. See [27]. ◻

Proof. If \({\mathcal{M}}\) satisfies \((M3)'\), by [26], we have \(j/m_j\to 0\), \(j\to\infty\), where \(m_j = M_j/M_{j-1}\). Hence, as in the beginning of the proof of [26], for every \(H>0\), we have \[\frac{j!}{H^j M_j} = \frac{1}{Hm_1} \cdot \frac{2}{Hm_2}\cdots \frac{j}{Hm_j} \to 0,\] which implies our claim. ◻

Proof. First, the fact that \({\{m_j\}_{j\in\mathbb{N}_0}}\) is increasing is equivalent to condition \((M1)\). Indeed, we have \[m_{j+1}\geq m_j\;\Leftrightarrow\;\frac{M_{j+1}}{M_j}\geq \frac{M_j}{M_{j-1}}\;\Leftrightarrow\;M_{j-1}M_{j+1}\geq M_j^2,\] for every \(j\geq 1\), and the case \(j=0\) is trivial. Also, since \(m_1=1\), it follows \(m_j\geq 1\) for all \(j\geq 1\). Then, for every \(j\geq 1\) we have \[M_j = \frac{M_1}{M_0}\cdot\frac{M_2}{M_1}\cdots\frac{M_j}{M_{j-1}} = m_1m_2\cdots m_j \leq m_j^j,\] and the case \(j=0\) is trivial. ◻

Proof. See [28]. ◻

Proof. See [31]. ◻

Proof. See [33]. ◻

Proof. Let \(f\in{\mathcal{S}}_{\{{\mathcal{M}}\}}({\mathbb{R}}^n)\) and \(C_f,h_f>0\) be such that \[\sup_{x\in{\mathbb{R}}^n}|x^\beta\partial_x^\alpha f(x)|\leq C_fh_f^{|\alpha|+|\beta|}M_{|\alpha|}M_{|\beta|},\quad\forall \alpha,\beta\in\mathbb{N}_0^n.\]

Then, given \(\alpha\in\mathbb{N}_0^n\) and \(N\in\mathbb{N}_0\), we have \[\begin{align} |\langle x\rangle^{2N}\partial_x^\alpha f(x)| & =|(1+|x|^2)^N\partial_x^\alpha f(x)| &&\leq\sum_{\ell=0}^N\binom{N}{\ell}|x|^{2\ell}|\partial_x^\alpha f(x)|\\ & \leq\sum_{\ell=0}^N\binom{N}{\ell}\sum_{|\beta|=2\ell}\frac{(2\ell!)}{\beta!}|x^\beta\partial_x^\alpha f(x)| &&\leq\sum_{\ell=0}^N\binom{N}{\ell}\sum_{|\beta|=2\ell}\frac{(2\ell)!}{\beta!}C_fh_f^{|\alpha|+|\beta|} M_{|\alpha|}M_{|\beta|}\\ & \leq C_f h_0^{2N} h_f^{|\alpha|}M_{|\alpha|}M_{2N}\sum_{\ell=0}^N\binom{N}{\ell}\sum_{|\beta|=2\ell}\frac{(2\ell)!}{\beta!}\\ & \leq C_f h_0^{2N} h_f^{|\alpha|}n^{2N}M_{|\alpha|}M_{2N} \sum_{\ell=0}^N\binom{N}{\ell}\sum_{|\beta|=2\ell} 1\\ & \leq C h_f^{|\alpha|}M_{|\alpha|}, \end{align}\] where \(h_0=\min\{h_f,1\}\) and \(C= C_f2^n(8h_0^2n^2)^N M_{2N}\). Also, we used the estimates and identities (1.2.1), (1.2.3), (1.2.6), and (1.2.7) from [25], and the fact that the number of multi-indices \(\beta \in \mathbb{N}_0^n\) such that \(|\beta|=2\ell\) is bounded by \(2^{n+2\ell}\) (see the comment in the end of page 11 of [25]). The projective case is analogous. The proof is complete. ◻

Proof. The proof follows the same lines of the one of Lemma 2, employing the Leibniz rule. ◻

Proof. See [35]. ◻

Proof. Let \(\chi\in{\mathcal{D}}_{[{\mathcal{M}}]}({\mathbb{R}})\) be a non-negative function such that \(\mathrm{supp}(\chi)\subset [-1,1]\) and \(\chi\equiv 1\) in \(\left[-\frac{1}{2},\frac{1}{2}\right]\). Then, consider \[\omega_0(t) = \int_{-1}^1 \omega_{\mathcal{M}}(t+1-s)\chi(s)\,\mathrm{d}s.\]

Applying the logarithm on Proposition 11, we obtain \[\omega_{\mathcal{M}}(t_1)+\omega_{\mathcal{M}}(t_2) \leq \omega_{\mathcal{M}}(H(t_1+t_2))+\log(A)\] and \[\omega_{\mathcal{M}}(t_1+t_2) \leq \omega_{\mathcal{M}}(2t_1)+\omega_{\mathcal{M}}(2t_2),\] for all \(t_1,t_2\geq 0\), where \(A,H\geq 1\) are given by property \((M2)'\).

(i) Given \(t_1,t_2\geq 0\), we have \[\begin{align} \omega_0(t_1+t_2) & = \int_{-1}^1 \omega_{\mathcal{M}}(t_1+t_2+1-s)\chi(s)\,\mathrm{d}s\\ & = \int_{-1}^1 \omega_{\mathcal{M}}\left(t_1+\frac{1-s}{2}+t_2 + \frac{1-s}{2}\right)\chi(s)\,\mathrm{d}s\\ & \leq \int_{-1}^1\omega_{\mathcal{M}}(2t_1+1-s)\chi(s)\,\mathrm{d}s + \int_{-1}^1\omega_{\mathcal{M}}(2t_2+1-s)\chi(s)\,\mathrm{d}s\\ & = \omega_0(2t_1)+\omega_0(2t_2). \end{align}\]

(ii) Given \(t_1,t_2\geq 0\), we have \[\begin{align} \omega_0(t_1)+\omega_0(t_2) & = \int_{-1}^1 \left( \omega_{\mathcal{M}}(t_1+1-s)+\omega_{\mathcal{M}}(t_2+1-s)\right)\chi(s)\,\mathrm{d}s\\ & \leq \int_{-1}^1 \omega_{\mathcal{M}}(H(t_1+t_2+2-2s))\chi(s)\,\mathrm{d}s + \log(A)\int_{-1}^1\chi(s)\,\mathrm{d}s\\ & = \int_{-1}^1 \omega_{\mathcal{M}}\left(H(t_1+t_2)+\frac{1-s}{2}+\frac{(2H-1)(1-s)}{2}\right)\chi(s)\,\mathrm{d}s + C_0\\ & \leq \omega_0(2H(t_1+t_2))+C, \end{align}\] where \[C_0 = \log(A)\int_{-1}^1\chi(s)\,\mathrm{d}s \quad\text{and}\quad C = \int_{-1}^{1}\omega_{\mathcal{M}}((2H-1)(1-s))\chi(s)\,\mathrm{d}s + C_0.\]

(iii) We have \[\omega_{\mathcal{M}}(t) \leq \omega_{\mathcal{M}}(t+1-s) = \omega_{\mathcal{M}}(t+1-s)\chi(s) + \omega_{\mathcal{M}}(t+1-s)(1-\chi(s)),\] for all \(s\in[-1,1]\). Then \[\begin{align} \omega_{\mathcal{M}}(t) & = \int_{-\frac{1}{2}}^{\frac{1}{2}}\omega_{\mathcal{M}}(t)\,\mathrm{d}s\\ & \leq \int_{-\frac{1}{2}}^{\frac{1}{2}}\omega_{\mathcal{M}}(t+1-s)\chi(s)\,\mathrm{d}s + \int_{-\frac{1}{2}}^{\frac{1}{2}}\omega_{\mathcal{M}}(t+1-s)(1-\chi(s))\,\mathrm{d}s\\ & \leq \omega_0(t), \end{align}\] for all \(t\geq 0\), noticing that \(1-\chi(s)\) vanishes for \(|s|\leq \frac{1}{2}\) and \(\omega_{\mathcal{M}}(t+1-s)\chi(s)\geq 0\).

On the other hand, by (ii) and property \((W1)\) of \(\omega_{\mathcal{M}}\), we can obtain \(C_1,C_2>0\) such that \[\begin{align} \omega_0(t) & = \int_{-1}^1 \omega_{\mathcal{M}}(t+1-s)\chi(s)\,\mathrm{d}s \leq \omega_{\mathcal{M}}(t+2)\int_{-1}^1\chi(s)\,\mathrm{d}(s)\\ & \leq \left(\omega_{\mathcal{M}}(2t)+\omega_{\mathcal{M}}(4)\right)\int_{-1}^1 \chi(s)\,\mathrm{d}s \leq C_1\omega_{\mathcal{M}}(t)+C_2, \end{align}\] for all \(t\geq 0\). Hence \(\omega_0\asymp \omega_{\mathcal{M}}\).

(iv) Finally, by the properties of the convolution, condition \((W2)\), and the fact that \(\chi\in{\mathcal{E}}_{[{\mathcal{M}}]}({\mathbb{R}})\), we have that there exist \(C_1,h>0\) (respectively, for every \(h>0\), there exists \(C_1>0\)) such that \[\begin{align} |\partial_t^k\omega_0(t)|\leq\;& \int_{-1}^1 \omega_{\mathcal{M}}(t+1-s)|\partial_s^k\chi(s)|\,\mathrm{d}s \leq \omega_{\mathcal{M}}(t+1)Ch^kM_k\\ & \leq C_1h^{k}M_k(\omega_{\mathcal{M}}(t)+\omega_{\mathcal{M}}(1)) \leq Ch^kM_k \omega_0(t), \end{align}\] for \(t\geq\delta\) and \(C>0\) not depending on \(k\). Here, we used \(\omega_{\mathcal{M}}\asymp\omega_0\). Finally, since \(\omega_0(t)\) is bounded on compact subsets (it is smooth by the regularizing property of the convolution), the previous estimate yields \(\omega_0\in{\mathcal{E}}_{[{\mathcal{M}}]}((0,+\infty))\). The proof is complete. ◻

Proof. See [37]. ◻

Proof. See [38]. ◻

Proof. Let \(\mathscr{B}\subset{\mathcal{S}}_{\{{\mathcal{M}}\}}({\mathbb{R}}^n)\) be a bounded subset, that is, there exist \(C,h>0\) such that \[\sup_{x\in{\mathbb{R}}^n} |x^\beta\partial_x^\alpha u(x)| \leq Ch^{|\alpha|+|\beta|}M_{|\alpha|}M_{|\beta|},\] for all \(\alpha,\beta\in{\mathbb{N}}_0^n\) and \(u\in\mathscr{B}\). Then, given \(u\in \mathscr{B}\), Leibniz rule and integration by parts yield \[\begin{align} x^\beta D_x^\alpha[\mathrm{Op}(p)u](x) & = x^\beta\sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma}\int_{{\mathbb{R}}^n} (D_x^{\gamma}e^{i\xi\cdot x})(D_x^{\alpha-\gamma}p(x,\xi))\widehat{u}(\xi)\,\mathrm{d}\bar{}\xi\\ & = \sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma}\int_{{\mathbb{R}}^n} (D_\xi^{\beta}e^{i\xi\cdot x})\xi^{\gamma}(D_x^{\alpha-\gamma}p(x,\xi))\widehat{u}(\xi)\,\mathrm{d}\bar{}\xi\\ & = (-1)^{|\alpha|}\sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma} \sum_{\delta\leq\beta}\binom{\beta}{\delta}\int_{{\mathbb{R}}^n} e^{i\xi\cdot x} (D_x^{\alpha-\gamma}D_\xi^{\beta-\delta}p(x,\xi)) D_\xi^{\delta}(\xi^\gamma\widehat{u}(\xi))\,\mathrm{d}\bar{}\xi. \end{align}\]

Now, notice that, for every \(N\in{\mathbb{N}}_0\), we have \[\begin{align} \int_{{\mathbb{R}}^n} e^{i\xi\cdot x} (D_x^{\alpha-\gamma}D_\xi^{\beta-\delta}&p(x,\xi)) D_\xi^{\delta}(\xi^\gamma\widehat{u}(\xi))\,\mathrm{d}\bar{}\xi\\ =\;& \langle x\rangle^{-2N}\int_{{\mathbb{R}}^n} e^{i\xi\cdot x}(1-\Delta_\xi)^N[(D_x^{\alpha-\gamma}D_\xi^{\beta-\delta}p(x,\xi)) D_\xi^{\delta}(\xi^\gamma\widehat{u}(\xi))]\,\mathrm{d}\bar{}\xi. \end{align}\]

Given \(\ell\in{\mathbb{N}}_0\), we set \[\Pi(\ell)=\{\lambda=(\lambda_1,\dots,\lambda_n)\in{\mathbb{N}}_0^n\,:\, \lambda_1+\cdots+\lambda_n = \ell\}.\] By the Lebniz rule and the multinomial theorem, given \(f,g\in C^\infty({\mathbb{R}}^n)\), we have \[\begin{align} (1-\Delta_\xi)^N(fg) & = \sum_{\ell=0}^N\binom{N}{\ell}(-1)^\ell\Delta_\xi^\ell(fg) = \sum_{\ell=0}^N\binom{N}{\ell}(-1)^\ell (\partial_{\xi_1}^2+\cdots+\partial_{\xi_n}^2)^\ell(fg)\\ &= \sum_{\ell=0}^N\binom{N}{\ell}(-1)^\ell \sum_{\lambda\in\Pi(\ell)}\dfrac{\ell!}{\lambda_1!\cdots\lambda_n!}\partial_\xi^{2\lambda}(fg)\\ & = \sum_{\ell=0}^N\binom{N}{\ell}(-1)^\ell \sum_{\lambda\in\Pi(\ell)}\dfrac{\ell!}{\lambda_1!\cdots\lambda_n!} \sum_{\nu\leq 2\lambda}\binom{2\lambda}{\nu} (\partial_{\xi}^{\nu}f)(\partial_{\xi}^{2\lambda-\nu} g). \end{align}\] Hence, for any \(N\in{\mathbb{N}}_0\), we obtain \[x^\beta\partial_x^\alpha[\mathrm{Op}(p)u](x) = \sum \int_{{\mathbb{R}}^n} e^{i\xi\cdot x}\langle\xi\rangle^{-2N} (D_x^{\alpha-\gamma}\partial_\xi^{2\lambda-\nu}D_\xi^{\beta-\delta}p(x,\xi))(\langle\xi\rangle^{2N}\partial_\xi^\nu D_\xi^\delta (\xi^\gamma\widehat{u}(\xi)))\,\mathrm{d}\bar{}\xi,\] where the summation symbol above denotes \[(-1)^{|\alpha|}\sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma} \sum_{\delta\leq\beta}\binom{\beta}{\delta}\sum_{\ell=0}^N\binom{N}{\ell}(-1)^\ell \sum_{\lambda\in\Pi(\ell)}\dfrac{\ell!}{\lambda_1!\cdots\lambda_n!} \sum_{\nu\leq 2\lambda}\binom{2\lambda}{\nu}.\] By the multinomial theorem, one can easily check that \[\label{bound95sum95binom} \sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma} \sum_{\delta\leq\beta}\binom{\beta}{\delta}\sum_{\ell=0}^N\binom{N}{\ell} \sum_{\lambda\in\Pi(\ell)}\dfrac{\ell!}{\lambda_1!\cdots\lambda_n!} \sum_{\nu\leq 2\lambda}\binom{2\lambda}{\nu} \leq (8n)^N 2^{|\alpha|+|\beta|}.\tag{6}\] Since \(p\in {{\mathrm{SG}}}_{\{{\mathcal{A}},{\mathcal{B}}\}}^{m,\mu}({\mathbb{R}}^n)\) and \({\mathcal{A}},{\mathcal{B}}\preceq{\mathcal{M}}\), there exist \(C_p,h_p>0\) such that \[\label{est95p} |D_x^{\alpha-\gamma}\partial_\xi^{2\lambda-\nu}D_\xi^{\beta-\delta}p(x,\xi)| \leq C_ph_p^{|\alpha-\gamma|+|\beta-\delta|}h_p^{2N}M_{|\alpha-\gamma|}M_{|\beta-\delta|}M_{2N}\langle\xi\rangle^{\mu}\langle x\rangle^{m}.\tag{7}\] Combining Lemmas 2 and 3, we obtain constants \(C_N,h_N>0\) depending on \(N\) such that \[\label{est95xi95u} |\langle\xi\rangle^{2N}\partial_\xi^\nu D_\xi^\delta (\xi^\gamma\widehat{u}(\xi))| \leq C_Nh_N^{|\gamma|+|\delta|}M_{|\gamma|}M_{|\delta|}.\tag{8}\] It follows, choosing \(N\in{\mathbb{N}}_0\) such that \(2N\geq \max\{\mu+n,m\}\), \(\langle x\rangle^{m-2N}\leq 1\) and \[\label{const95int} \tilde{C} = \int_{{\mathbb{R}}^n}\langle\xi\rangle^{\mu-2N}\,\mathrm{d}\bar{}\xi<+\infty.\tag{9}\]

Then, by 6 , 7 , and 8 , we obtain \[|x^\beta\partial_x^\alpha u(x)| \leq C_0h_0^{|\alpha|+|\beta|}M_{|\alpha|}M_{|\beta|},\] for all \(x\in{\mathbb{R}}^n\), \(\alpha,\beta\in{\mathbb{N}}_0^n\), and \(u\in\mathscr{B}\), where the constants \[C_0 = \tilde{C}C_NC_p(8h_p^{2}n)^NM_{2N}\quad\text{and}\quad h_0 = 2\max\{1,h_N,h_p\}\] do not depend on \(u\in\mathscr{B}\). Hence, \(\mathrm{Op}(p)\) is bounded from \({\mathcal{S}}_{\{{\mathcal{M}}\}}({\mathbb{R}}^n)\) to itself. The projective case is analogous. ◻

Proof. See [33]. ◻

Proof. For each \(N\in{\mathbb{N}}_0\), set \[K_N(x,y)=\int_{R^n}e^{i\xi\cdot(x-y)}p(x,\xi)\psi_N(\xi)\,\mathrm{d}\bar{}\xi,\] where the \(\psi_N\) are given by Lemma 7. In this case, we obtain \[K_p(x,y) = \sum_{N\in{\mathbb{N}}_0}K_N(x,y).\]

Let \(k\in(0,1)\) and \((x,y)\in\overline{\Omega}_k\) and consider \(\ell\in\{1,\dots,n\}\) such that \[\label{x95ell95y95ell} |x_\ell- y_\ell| \geq \frac{k}{n\langle x\rangle},\tag{10}\] where \(x_\ell\) and \(y_\ell\) denote the \(\ell\)-th coordinate of \(x\) and \(y\), respectively. Then, by the Leibniz rule and integration by parts, we obtain \[\begin{align} D_x^\beta D_y^\gamma K_N(x,y) & = (-1)^{|\gamma|}\sum_{\delta\leq\beta}\binom{\beta}{\delta}\int_{{\mathbb{R}}^n}e^{i\xi\cdot (x-y)}\xi^{\delta+\gamma}\psi_N(\xi) D_x^{\beta-\delta}p(x,\xi)\,\mathrm{d}\bar{}\xi\\ & = (-1)^{|\gamma|+N}(x_\ell-y_\ell)^{-N}\sum_{\delta\leq\beta}\binom{\beta}{\delta}\int_{{\mathbb{R}}^n}e^{i\xi\cdot (x-y)}D_{\xi_\ell}^N\left(\xi^{\delta+\gamma}\psi_N(\xi) D_x^{\beta-\delta}p(x,\xi)\right)\,\mathrm{d}\bar{}\xi. \end{align}\]

Given \(\rho>0\), we define the operator \[\mathcal{L}_\rho = \frac{1}{\lambda_\rho(\langle x-y\rangle)}\sum_{j\in{\mathbb{N}}_0}\frac{\rho^{2j}}{M_{j}^2}(1-\Delta_\xi)^j,\] where \[\lambda_\rho(t) = \sum_{j\in{\mathbb{N}}_0}\frac{\rho^{2j}}{M_{2j}} t^{2j},\quad t > 0.\]

By 4 , we have \[\label{est95lambda95rho} e^{\omega_{{\mathcal{M}}}(\rho t)} \leq \lambda_\rho(t),\tag{11}\] for all \(\rho,t>0\).

It is not difficult to check that \(\mathcal{L}_\rho e^{i\xi(x-y)} = e^{i\xi(x-y)}\). Hence, integrating by parts, we obtain \[\begin{align} D_x^\beta D_y^\gamma K_N(x,y) =\;& \frac{(-1)^{|\gamma|+N}(x_\ell-y_\ell)^{-N}}{\lambda_\rho(x,y)}\sum_{\delta\leq\beta}\binom{\beta}{\delta} \sum_{j\in{\mathbb{N}}_0}\frac{\rho^{2j}}{M_j^2}\\ & \times \int_{{\mathbb{R}}^n}e^{i\xi\cdot (x-y)}(1-\Delta_\xi)^j D_{\xi_\ell}^N\left(\xi^{\delta+\gamma}\psi_N(\xi) D_x^{\beta-\delta}p(x,\xi)\right)\,\mathrm{d}\bar{}\xi. \end{align}\]

Proceeding as in the proof of Theorem 17, we can apply the binomial and multinomial theorems to expand \((1-\Delta_\xi)^j\), which yields \[\begin{align} \left| (1-\Delta_\xi)^j D_{\xi_\ell}^N\left(\xi^{\delta+\gamma}\psi_N(\xi) D_x^{\beta-\delta}p(x,\xi)\right) \right| \leq\;& \sum_{j'=0}^j\binom{j}{j'}\sum_{\lambda\in\Pi(j')}\frac{j'!}{\lambda_1!\cdots\lambda_n!}\sum_{\nu\leq 2\lambda}\binom{2\lambda}{\nu}\sum_{N'=0}^N\binom{N}{N'}\\ & \times |\partial_\xi^\nu\partial_{\xi_\ell}^{N'}(\xi^{\gamma+\delta}\psi_N(\xi))||\partial_\xi^{2\lambda-\nu}\partial_{\xi_\ell}^{N-N'}\partial_{x}^{\beta-\delta}p(x,\xi)|. \end{align}\]

Now, by Lemmas 3 and 7 and the fact that \(\mathcal{D}_{[{\mathcal{M}}]}({\mathbb{R}}^n)\subset{\mathcal{S}}_{[{\mathcal{M}}]}({\mathbb{R}}^n)\), given \(R>0\), there exist \(C_\psi,h_\psi>0\) (respectively, for every \(h_\psi>0\), there exists \(C_\psi>0\)) such that \[|\partial_\xi^\nu\partial_{\xi_\ell}^{N'}(\xi^{\gamma+\delta}\psi_N(\xi))| \leq C_\psi h_\psi^{|\gamma|+|\delta|} M_{|\nu|}M_{N'}M_{|\gamma|+|\delta|}\left(\frac{h_\psi}{Rm_N}\right)^{|\nu|+N'}.\]

Also, since \(p\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m,\mu}({\mathbb{R}}^n)\) and \({\mathcal{A}},{\mathcal{B}}\preceq {\mathcal{M}}\), there exist \(C_p,h_p>0\) (respectively, for every \(h_p>0\), there exists \(C_p>0\)) such that \[\begin{align} |\partial_\xi^{2\lambda-\nu}\partial_{\xi_\ell}^{N-N'}\partial_{x}^{\beta-\delta}p(x,\xi)| \leq\;& C_ph_p^{|2\lambda|-|\nu|+N-N'+|\beta|-|\delta|}M_{|2\lambda|-|\nu|}M_{N-N'}M_{|\beta-\delta|}\\ & \times \langle\xi\rangle^{\mu-|2\lambda-\nu|-N+N'} \langle x\rangle^{m-|\beta-\delta|}. \end{align}\]

We have the following bound for the binomial and multinomial sums: \[\sum_{j'=0}^j\binom{j}{j'}\sum_{\lambda\in\Pi(j')}\frac{j'!}{\lambda_1!\cdots\lambda_n!}\sum_{\nu\leq 2\lambda}\binom{2\lambda}{\nu}\sum_{N'=0}^N\binom{N}{N'} \leq 8^N(2n)^j.\]

Combining the previous estimates, we obtain constants \(C_1,h_1>0\) (respectively, for every \(h_1>0\), we obtain \(C_1>0\)) such that \[\begin{align} \left| (1-\Delta_\xi)^j D_{\xi_\ell}^N\left(\xi^{\delta+\gamma}\psi_N(\xi) D_x^{\beta-\delta}p(x,\xi)\right) \right| \leq\;& C_1h_1^{2j+N+|\beta|+|\gamma|}M_{j}^2M_NM_{|\beta|}M_{|\gamma|}\left(\frac{1}{Rm_N}\right)^{|\nu|-N'}\\ & \times \langle\xi\rangle^{\mu - |2\lambda-\nu|-(N-N')} \langle x\rangle^{m}. \end{align}\]

Now, since \(2Rm_N < \langle\xi\rangle < 3Rm_{N+1}\) on \(\mathrm{supp}(\psi_N)\), if \(\mu> 0\), we obtain \[\langle\xi\rangle^{\mu - |2\lambda-\nu|-(N-N')} \leq (3Rm_{N+1})^{\mu}\left(\frac{1}{2Rm_N}\right)^{|2\lambda-\nu|+(N-N')},\] and \[(3Rm_{N+1})^{\mu} \leq (3R)^{\mu}\tilde{C}(N+1)^{\mu} \leq (3R)^{\mu}\tilde{C}_{\mu}e^N,\] where \(\tilde{C}>0\) do not depend on \(N\) and \(\tilde{C}_{\mu}\) depends only on \(\tilde{C}\) and \(\mu\).

If \(\mu\leq 0\), \((3Rm_{N+1})^{\mu}\) is bounded by a constant \(\tilde{C}>0\) independent of \(N\). Hence, we obtain new constants \(C_2,h_2>0\) (respectively, for every \(h_2>0\), there exists \(C_2>0\)) such that \[\begin{align} \left| (1-\Delta_\xi)^j D_{\xi_\ell}^N\left(\xi^{\delta+\gamma}\psi_N(\xi) D_x^{\beta-\delta}p(x,\xi)\right) \right| \leq\;& C_2 h_2^{|\beta|+|\gamma|}M_{|\beta|}M_{|\gamma|}h_2^{2j}M_{j}^2M_N \langle x\rangle^{m} \left(\frac{h_2}{Rm_N}\right)^{|2\lambda|+N}\\ \leq\;& C_2 h_2^{|\beta|+|\gamma|}M_{|\beta|}M_{|\gamma|}h_2^{2j}M_{j}^2 \langle x\rangle^{m} \left(\frac{h_2}{R}\right)^{|2\lambda|+N},\\ \leq\;& C_2 h_2^{|\beta|+|\gamma|}M_{|\beta|}M_{|\gamma|}h_2^{2j}M_{j}^2 \langle x\rangle^{m}, \end{align}\] for a sufficiently big \(R>0\), and we used Lemma 1 in the second inequality.

Hence, by the previous estimates and 10 and 11 , we obtain \(C,h>0\) (respectively, for every \(h>0\), we obtain \(C>0\)) such that \[\begin{align} |D_x^\beta D_y^\gamma K_N(x,y)| \leq\;& \frac{|x_\ell-y_\ell|^{-N}}{\lambda_\rho(\langle x-y\rangle)}\sum_{\delta\leq\beta}\binom{\beta}{\delta}\sum_{j\in{\mathbb{N}}_0}(\rho h_2)^{2j} C_0h_2^{|\beta|+|\gamma|}M_{|\beta|}M_{|\gamma|}\langle x\rangle^{m}\\ \leq & Ch^{|\beta|+|\gamma|}M_{|\beta|}M_{|\gamma|}\langle x\rangle^{m-N}e^{-\omega_{{\mathcal{M}}}(\rho\langle x-y\rangle)}, \end{align}\] where we choose \(\rho>0\) small enough such that the series \(\sum_{j\in{\mathbb{N}}_0}(\rho h_2)^{2j}\) is convergent.

Finally, since \((x,y)\in\overline{\Omega}_k\), we obtain \(C'>0\) such that \(\langle x-y\rangle \geq C'(\langle x\rangle+\langle y\rangle) \geq C'(|x|+|y|)\). Then, for such \((x,y)\) and the properties of \(\omega_{{\mathcal{M}}}\), we obtain \[\langle x\rangle^{m-N}e^{-\omega_{{\mathcal{M}}}(\rho\langle x-y\rangle)} \leq e^{-\omega_{{\mathcal{M}}}(\rho'|x|)}e^{-\omega_{{\mathcal{M}}}(\rho'|y|)},\] where \(\rho'>0\) is chosen in such a way that the exponential absorbs the polynomial \(\langle x\rangle^{m}\). Such \(\rho'\) can be obtained by Propositions 1 and 2 of [35]. The proof is complete. ◻

Proof. Let \(\varphi_1\in\mathcal{D}_{[{\mathcal{A}}]}({\mathbb{R}}^n)\) and \(\varphi_2\in\mathcal{D}_{[{\mathcal{B}}]}({\mathbb{R}}^n)\) be such that \(0\leq\varphi_\ell\leq 1\), \(\varphi_\ell(t)\equiv 1\) if \(\langle t\rangle \leq 2\), and \(\varphi_\ell(t)\equiv 0\) if \(\langle t\rangle \geq 3\), for \(\ell=1,2\). Set \(\chi(x,\xi)=1-\varphi_1(\xi)\varphi_2(x)\). Then, for \(j\in{\mathbb{N}}\) and \(R>0\), we define \[\chi_j(x,\xi)=\chi\left(\frac{x}{Rb_j},\frac{\xi}{Ra_j}\right),\] and put \(\chi_0(x,\xi)\equiv 1\). By definition, there exist \(C_\chi,h_\chi>0\) (respectively, for every \(h_\chi>0\), there exists \(C_\chi>0\)) such that \[\label{der95chi95j} \sup_{(x,\xi)\in{\mathbb{R}}^{2n}}|D_\xi^\alpha D_x^\beta \chi_j(x,\xi)| \leq \frac{C_\chi h_\chi^{|\alpha|+|\beta|}A_{|\alpha|}B_{|\beta|}}{R^{|\alpha|+|\beta|}a_j^{|\alpha|}b_j^{|\beta|}},\tag{12}\] for all \(\alpha,\beta\in{\mathbb{N}}_0^n\) and \(j\in{\mathbb{N}}_0\).

Let us show that \[\label{p95chi95pj} p(x,\xi) = \sum_{j\in{\mathbb{N}}_0}\chi_j(x,\xi)p_j(x,\xi)\tag{13}\] belongs to \({{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m,\mu}({\mathbb{R}}^n)\) for a sufficiently big \(R>0\), and \(p\sim \sum_{j\in{\mathbb{N}}_0}p_j\) in \({\mathrm{FS}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m,\mu}({\mathbb{R}}^n)\).

Given \(\alpha,\beta\in{\mathbb{N}}_0^n\), differentiating 13 yields \[D_\xi^\alpha D_\xi^\beta p(x,\xi) = \sum_{j\in{\mathbb{N}}_0} \sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma} \sum_{\delta\leq\beta}\binom{\beta}{\delta} D_\xi^{\alpha-\gamma}D_x^{\beta-\delta}p_j(x,\xi) D_\xi^\gamma D_x^\delta \chi_j(x,\xi).\]

Let us consider the inductive case. Since \(\sum_{j\in{\mathbb{N}}_0} p_j \in {\mathrm{FS}}_{\{{\mathcal{A}},{\mathcal{B}}\}}^{m,\mu}({\mathbb{R}}^n)\), there exist \(C,h>0\) such that \[|D_\xi^{\alpha-\gamma}D_x^{\beta-\delta}p_j(x,\xi)| \leq Ch^{|\alpha-\gamma|+|\beta-\delta|+2j}A_{|\alpha-\gamma|}B_{|\beta-\delta|}A_jB_j\langle\xi\rangle^{m_\xi-|\alpha-\gamma|-j}\langle x\rangle^{m-|\beta-\delta|-j},\] for \((x,\xi)\in Q_{B(j)}^c\). Then, taking \(R>B\), we obtain \[\begin{align} |D_\xi^\alpha D_x^\beta p(x,\xi)| & \leq C\langle\xi\rangle^{\mu-|\alpha|}\langle x\rangle^{m-|\beta|}\sum_{j\in{\mathbb{N}}_0}\sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma} \sum_{\delta\leq\beta}\binom{\beta}{\delta} h^{|\alpha-\gamma|+|\beta-\delta|+2j}\langle\xi\rangle^{|\gamma|-j}\langle x\rangle^{|\delta|-j}\\ & \times A_{|\alpha-\gamma|}B_{|\beta-\delta|}A_jB_j |D_\xi^\gamma D_x^\delta\chi_j(x,\xi)|\\ & \leq Ch^{|\alpha|+|\beta|} A_{|\alpha|}B_{|\beta|} \langle\xi\rangle^{\mu-|\alpha|}\langle x\rangle^{m-|\beta|} \sum_{j\in{\mathbb{N}}_0} \frac{h^{2j}A_jB_j}{\langle\xi\rangle^{j}\langle x\rangle^{j}}|\chi_j(x,\xi)|\\ & + C \langle\xi\rangle^{\mu-|\alpha|}\langle x\rangle^{m-|\beta|}\sum_{j\in{\mathbb{N}}_0}\sum_{(\gamma,\delta)\neq(0,0)}\binom{\alpha}{\gamma}\binom{\beta}{\delta}h^{|\alpha-\gamma|+|\beta-\delta|+2j}A_{|\alpha-\gamma|}B_{|\beta-\delta|}\\ & \times \frac{\langle\xi\rangle^{|\gamma|}\langle x\rangle^{|\delta|} A_jB_j}{\langle\xi\rangle^j\langle x\rangle^j}|D_\xi^\gamma D_x^\delta\chi_j(x,\xi)|\\ & = S_1 + S_2. \end{align}\]

To estimate \(S_1\), notice that \(\langle\xi\rangle\langle x\rangle \geq R^2a_jb_j\) on \(\mathrm{supp}(\chi_j)\). Hence \[\sum_{j\in{\mathbb{N}}_0} \frac{h^{2j}A_jB_j}{\langle\xi\rangle^{j}\langle x\rangle^{j}}|\chi_j(x,\xi)| \leq \sum_{j\in{\mathbb{N}}_0} \frac{h^{2j}A_jB_j}{R^{2j}a_j^jb_j^j} \leq \sum_{j\in{\mathbb{N}}_0}\frac{h^{2j}}{R^{2j}} = C_{R} < +\infty,\] for a large enough \(R>0\), where we used Lemma 1. It follows \[S_1 \leq CC_Rh^{|\alpha|+|\beta|}A_{|\alpha|}B_{|\beta|}\langle\xi\rangle^{\mu-|\alpha|}\langle x\rangle^{m-|\beta|},\] for all \(\alpha,\beta\in{\mathbb{N}}_0^n\).

Now, for \(S_2\), since \((\gamma,\delta)\neq (0,0)\), observe that \(D_\xi^\delta D_x^\gamma\chi_j\equiv 0\) on \(Q_{3R(j)}^c\). For \((x,\xi)\in Q_{3R(j)}\), we have \(\langle\xi\rangle^{|\gamma|}\langle x\rangle^{|\delta|}\leq (3R)^{|\gamma|+|\delta|}a_j^{|\gamma|}b_j^{|\delta|}\). Hence, using 12 , we obtain \[\begin{align} & h^{|\alpha-\gamma|+|\beta-\delta|+2j}A_{|\alpha-\gamma|}B_{|\beta-\delta|} A_jB_j \frac{\langle\xi\rangle^{|\gamma|}\langle x\rangle^{|\delta|} }{\langle\xi\rangle^j\langle x\rangle^j}|D_\xi^\gamma D_x^\delta\chi_j(x,\xi)|\\ \leq\;& h^{|\alpha-\gamma|+|\beta-\delta|+2j} A_{|\alpha-\gamma|} B_{|\beta-\delta|}A_jB_j\frac{ (3R)^{|\gamma|+|\delta|}a_j^{|\gamma|}b_j^{|\delta|}}{R^{2j}a_j^j b_j^j} \frac{C_\chi h_\chi^{|\gamma|+|\delta|} A_{|\gamma|}B_{|\delta|}}{R^{|\gamma|+|\delta|}a_j^{|\gamma|}b_j^{|\delta|}}\\ \leq\;& C_\chi h_0^{|\alpha|+|\beta|}A_{|\alpha|}B_{|\beta|} \frac{h^{2j}}{R^{2j}}, \end{align}\] where \(h_0 = \max\{h,h_\chi,3\}\). Hence, we have \[S_2 \leq CC_\chi C_R (2h_0)^{|\alpha|+|\beta|}A_{|\alpha|}B_{|\beta|}\langle\xi\rangle^{\mu-|\alpha|}\langle x\rangle^{m-|\beta|},\] where the factor \(2\) multiplicating \(h_0\) appeared from the sum over binomial coefficients. Combining the estimates for \(S_1\) and \(S_2\), we conclude that \(p\in {{\mathrm{SG}}}_{\{{\mathcal{A}},{\mathcal{B}}\}}^{m,\mu}({\mathbb{R}}^n)\).

Now, notice that, for every \(N\in{\mathbb{N}}_0\), \[p(x,\xi)-\sum_{j<N}p_j(x,\xi) = \sum_{j\geq N}p_j(x,\xi)\chi_j(x,\xi),\] for \((x,\xi)\in Q_{3R(j)}^c\). Proceeding as above, we can show that \(p(x,\xi)-\sum_{j<N}p_j(x,\xi)\) satisfies the estimates of Definition 8, hence \(p\sim \sum_{j\in{\mathbb{N}}_0}p_j\) in \({\mathrm{FS}}_{\{{\mathcal{A}},{\mathcal{B}}\}}^{m,\mu}({\mathbb{R}}^n)\).

The projective case can be proved analogously by doing the suitable choices on the constants \(h\) and \(h_\chi\). ◻

Proof. See [33]. ◻

Proof. Since \(p\sim 0\), there exist \(C,h,B>0\) (respectively, there exists \(B>0\) and, for every \(h>0\), there exists \(C>0\)) such that, for all \(N\in{\mathbb{N}}_0\) and \((x,\xi)\in Q_{B(N)}^c\), we have \[|D_\xi^\alpha D_x^\beta p(x,\xi)| \leq Ch^{|\alpha|+|\beta|}M_{|\alpha|}M_{|\beta|}\langle\xi\rangle^{-|\alpha|}\langle x\rangle^{-|\beta|}\frac{M_N}{h^N}.\]

If \(h>1\), let \(\ell=h^{-1}\). For \(\langle y\rangle\geq Bm_N\), we have \[\frac{h^NM_N}{\langle y\rangle^N} \leq \frac{M_N}{\ell^N|y|^N}.\]

If \(h\leq 1\), choose any \(h_0>1\) and let \(\ell=h_0^{-1}\). In this case, for \(\langle y\rangle\geq Bm_N\), we have \[\frac{h^NM_N}{\langle y\rangle^N} \leq \frac{h_0^NM_N}{|y|^N} = \frac{M_N}{\ell^N|y|^N}.\]

Hence, applying Lemma 8, we obtain \(C',\varepsilon>0\) such that \[|D_\xi^\alpha D_x^\beta p(x,\xi)| \leq CC'h^{|\alpha|+|\beta|}M_{|\alpha|}M_{|\beta|} e^{-\omega_{{\mathcal{M}}}(\ell\varepsilon|x|)}e^{-\omega_{{\mathcal{M}}}(\ell\varepsilon|\xi|)},\] for all \((x,y)\in Q_B^c\) and \(\alpha,\beta\in{\mathbb{N}}_0\). By Proposition 12, we have \(p\in{\mathcal{S}}_{[{\mathcal{M}}]}({\mathbb{R}}^{2n})\), which implies that the kernel \(K_p\) of \({\mathrm{Op}}(p)\) also belongs to \({\mathcal{S}}_{[{\mathcal{M}}]}({\mathbb{R}}^{2n})\). We conclude that \({\mathrm{Op}}(p)\) is \([{\mathcal{M}}]\)-regularizing. ◻

Proof. The proof follows the same lines of [4], . ◻

Proof. We have \({}^t({}^t{\mathrm{Op}}(q))={\mathrm{Op}}(q)\) and \({}^t{\mathrm{Op}}(q)={\mathrm{Op}}(q_1)+R_1\), where \(R_1\) is \([{\mathcal{M}}]\)-regularizing and \(q_1\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{\mu',m'}({\mathbb{R}}^n)\) has asymptotic expansion \[\label{exp95q1} q_1(x,\xi) \sim \sum_{j\in{\mathbb{N}}_0}\sum_{|\alpha|=j}\frac{1}{\alpha!}\partial_\xi^\alpha D_x^\alpha q(x,-\xi),\tag{15}\] by Proposition 24. Moreover, \({\mathrm{Op}}(p){\mathrm{Op}}(q) = {\mathrm{Op}}(p)\,{}^t{\mathrm{Op}}(q_1)+R_2\), where \(R_2\) is \([{\mathcal{M}}]\)-regularizing. Notice that \[{}^t{\mathrm{Op}}(q)u(x) = \int_{{\mathbb{R}}^n} e^{i\xi\cdot x}\left(\int_{{\mathbb{R}}^n}e^{-i\xi\cdot y}q_1(y,-\xi)u(y)\,\mathrm{d}\bar{}y\right)\mathrm{d}\bar{}\xi,\] which implies \[\widehat{({}^t{\mathrm{Op}}(q_1)u)}(\xi) = \int_{{\mathbb{R}}^n}e^{-i\xi\cdot y}q_1(y,-\xi)u(y)\,\mathrm{d}\bar{}y,\] and \[{\mathrm{Op}}(p)\,{}^t{\mathrm{Op}}(q_1) = \int_{{\mathbb{R}}^n}\int_{{\mathbb{R}}^n} e^{i\xi\cdot(x-y)}p(x,\xi)q_2(y,\xi)u(y)\,\mathrm{d}\bar{}y\mathrm{d}\bar{}\xi,\] where \(q_2(x,\xi)=q_1(x,-\xi)\). By 15 , we obtain \[\label{exp95q2} q_2(x,\xi) \sim \sum_{j\in{\mathbb{N}}_0}\sum_{|\alpha|=j}\frac{(-1)^{|\alpha|}}{\alpha!}\partial_\xi^\alpha D_x^\alpha q(x,\xi).\tag{16}\]

In this case, \({\mathrm{Op}}(p)\,{}^t{\mathrm{Op}}(q_1)\) is an operator with amplitude \(a(x,y,\xi) = p(x,\xi)q_2(y,\xi)\in {\mathrm{AG}}_{[{\mathcal{A}},{\mathcal{B}}]}^{\mu+\mu',m,m'}({\mathbb{R}}^n)\). By Theorem 23, we have \({\mathrm{Op}}(p)\,{}^t{\mathrm{Op}}(q_1) = {\mathrm{Op}}(s)+R\), where \(R\) is \([{\mathcal{M}}]\)-regularizing and \(s\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{\mu+\mu',m+m'}({\mathbb{R}}^n)\) satisfies \[s(x,\xi) \sim \sum_{j\in{\mathbb{N}}_0}\sum_{|\alpha|=j}\frac{1}{\alpha!}\partial_\xi^\alpha\left( p(x,\xi) D_x^\alpha q(x,\xi)\right).\]

It only remains to show that \[\sum_{j\in{\mathbb{N}}_0}\sum_{|\alpha|=j}\frac{1}{\alpha!}\partial_\xi^\alpha\left( p(x,\xi) D_x^\alpha q_2(x,\xi)\right) \sim \sum_{j=0}^\infty\sum_{|\alpha|=j}\frac{1}{\alpha!}\partial_\xi^\alpha p(x,\xi)D_x^\alpha q(x,\xi).\]

By the Leibniz rule and 16 , we have \[\begin{align} \sum_{\alpha}\frac{1}{\alpha!}\partial_\xi^\alpha\left( p(x,\xi) D_x^\alpha q_2(x,\xi)\right) & \sim \sum_{\alpha,\beta}\frac{1}{\alpha!\beta!}\partial_\xi^\alpha\left( p(x,\xi) \partial_\xi^\beta D_x^{\alpha+\beta}q(x,\xi)\right)\\ & \sim \sum_{\alpha,\beta}\frac{(-1)^{|\beta|}}{\alpha!\beta!}\sum_{\gamma+\delta=\alpha}\frac{\alpha!}{\gamma!\delta!}\partial_\xi^{\gamma}p(x,\xi)\partial_\xi^{\beta+\delta}D_x^{\beta+\gamma+\delta}q(x,\xi)\\ & \sim \sum_{\gamma,\nu}\left(\sum_{\beta+\delta=\nu}\frac{(-1)^{|\beta|}}{\beta!\delta!}\right) \frac{1}{\gamma!}\partial_\xi^{\gamma}p(x,\xi)\partial_\xi^{\beta+\delta}D_x^{\beta+\gamma+\delta}q(x,\xi). \end{align}\]

By the binomial formula, with \(\boldsymbol{e}=(1,\dots,1)\in{\mathbb{R}}^n\), we have that \[\sum_{\beta+\delta=\nu}\frac{(-1)^{|\beta|}}{\beta!\delta!} = \frac{1}{\nu!}\sum_{\beta+\delta=\nu}\frac{\nu!}{\beta!\delta!}(-1)^{|\beta|}\boldsymbol{e}^\beta \boldsymbol{e}^\delta = \frac{1}{\nu!} (\boldsymbol{e}-\boldsymbol{e})^{\nu}\] vanishes for \(\nu\not=0\). We conclude that the only non-vanishing terms in the asymptotic expansion are those with \(\nu=0\). The proof is complete. ◻

Proof. Let \(\alpha',\beta',\gamma'\in\mathbb{N}_0^n\). Then, there exist \(C_0,h_0>0\) such that \[\begin{align} |\partial_\xi^{\alpha'}\partial_x^{\beta'}\partial_y^{\gamma'} \partial_\xi^{\alpha}\partial_x^{\beta}\partial_y^{\gamma} p(x,y,\xi)| \leq\;& C_0h_0^{|\alpha|+|\beta|+|\gamma|}h^{|\alpha'|+|\beta'|+|\gamma'|} A_{|\alpha|+|\alpha'|}B_{|\beta|+|\beta'|}B_{|\gamma|+|\gamma'|} \\ & \times \langle\xi\rangle^{\kappa(|\beta|+|\gamma|+|\beta'|+|\gamma'|)} \langle (x,y)\rangle^{\lambda(|\alpha|+|\alpha'|)} \\ \leq &\;\left( C_0 (2h_0)^{|\alpha|+|\beta|+|\gamma|} A_{|\alpha|} B_{|\beta|} B_{|\gamma|} \right) (2h_0)^{|\alpha'|+|\beta'|+|\gamma'|}A_{|\alpha|'} B_{|\beta'|} B_{|\gamma'|}\\ & \times \langle\xi\rangle^{\kappa(|\beta|+|\gamma|+|\beta'|+|\gamma'|)} \langle (x,y)\rangle^{\lambda(|\alpha|+|\alpha'|)}. \end{align}\] Then, to prove the desired estimates, it is enough to take \(C = C_0(2h_0)^{|\alpha|+|\beta|+|\gamma|}A_{|\alpha|} B_{|\beta|} B_{|\gamma|}\), \(h=2h_0\), \(\kappa'(j)=\kappa(|\beta|+|\gamma|+j)\), and \(\lambda'(j)=\lambda(|\alpha|+j)\). ◻

Proof. First, since \(q\) is a polynomial in \((x,y,\xi)\), we can find a constant \(C_q>0\) such that \[|\partial_\xi^{\alpha'}\partial_x^{\beta'}\partial_y^{\gamma'} q(x,y,\xi)|\leq C_q\langle\xi\rangle^{d_1}\langle(x,y)\rangle^{d_2},\] for every \(\alpha',\beta',\gamma'\in\mathbb{N}_0^n\) and \(x,y,\xi\in\mathbb{R}^n\), where \(d_1,d_2\in\mathbb{N}_0\) are the \(\xi\)-degree and the \((x,y)\)-degree of \(q\), respectively.

Now, let \(\alpha,\beta,\gamma\in\mathbb{N}_0^n\). Then, \[\begin{align} |\partial_\xi^{\alpha} \partial_x^{\beta} \partial_y^{\gamma} &(pq)(x,y,\xi)| \leq \\ &\leq \sum_{\alpha'\leq\alpha}\binom{\alpha}{\alpha'} \sum_{\beta'\leq\beta}\binom{\beta}{\beta'} \sum_{\gamma'\leq\gamma}\binom{\gamma}{\gamma'} |\partial_\xi^{\alpha'}\partial_x^{\beta'}\partial_y^{\gamma'} q(x,y,\xi)||\partial_\xi^{\alpha-\alpha'}\partial_x^{\beta-\beta'}\partial_y^{\gamma-\gamma'} p(x,y,\xi)|\\ & \leq \sum_{\alpha'\leq\alpha}\binom{\alpha}{\alpha'} \sum_{\beta'\leq\beta}\binom{\beta}{\beta'} \sum_{\gamma'\leq\gamma}\binom{\gamma}{\gamma'} C_q\langle\xi\rangle^{d_1}\langle(x,y)\rangle^{d_2} Ch^{|\alpha|+|\beta|+|\gamma|} A_{|\alpha|}B_{|\beta|}B_{|\gamma|} \\ & \times \langle \xi\rangle^{\kappa(|\beta|+|\gamma|)}\langle(x,y)\rangle^{\lambda(|\alpha|)}\\ & \leq CC_q(2h)^{|\alpha|+|\beta|+|\gamma|} A_{|\alpha|}B_{|\beta|}B_{|\gamma|} \langle \xi\rangle^{d_1+\kappa(|\beta|+|\gamma|)}\langle(x,y)\rangle^{d_2+\lambda(|\alpha|)}, \end{align}\] where the term \(2^{|\alpha|+|\beta|+|\gamma|}\) is a consequence of the sums over binomial coefficients. Taking \(\kappa'(j)=N+\kappa(j)\) and \(\lambda'(j)=M+\lambda(j)\), we conclude the proof. ◻

Proof. Let \(u \in {\mathcal{S}}_{[{\mathcal{M}}]}({\mathbb{R}}^n)\) and \(N_1,N_2\in\mathbb{N}_0\). Integrating by parts, we obtain that \([\mathrm{Op}(p)u](x)\) equals \[\label{I95x95xi} \langle\xi\rangle^{-2N_1} \langle x-y\rangle^{-2N_2} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} e^{i\xi\cdot(x-y)} (1-\Delta_\xi)^{N_1} (1-\Delta_y)^{N_2} (p(x,y,\xi)u(y))\,\mathrm{d}\bar{}y\,\mathrm{d}\bar{}\xi.\tag{17}\]

By the Leibniz rule, \((1-\Delta_\xi)^{N_1}(1-\Delta_y)^{N_2}(p(x,y,\xi)u(y))\) is a linear combination of terms of the form \[(1-\Delta_\xi)^{N_1}\partial_y^{\gamma_1}p(x,y,\xi)\partial_y^{\gamma_2}u(y),\quad |\gamma_1+\gamma_2|\leq 2N_2.\] For such terms, given \(N_3\in\mathbb{N}_0\), we have \[\begin{align} |(1-\Delta_\xi)^{N_1}\partial_y^{\gamma_1}p(x,y,\xi)\partial_y^{\gamma_2}u(y)| & = \langle y\rangle^{-2N_3}|(1-\Delta_\xi)^{N_1}\partial_y^{\gamma_1}p(x,y,\xi)\langle y\rangle^{2N_3}\partial_y^{\gamma_2}u(y)|\\ & \leq Ch^{2N_1+2N_2+2N_3} A_{2N_1}B_{2N_2} B_{2N_3}\\ & \times \langle\xi\rangle^{\kappa(2N_2)}\langle (x,y)\rangle^{\lambda(2N_1)}\langle y\rangle^{-N_3}, \end{align}\] where we used Lemma 2, the fact that \(|\gamma_1|\leq 2N_2\) and that the function \(\kappa\) in non-decreasing. Then, for each \(N_3\in\mathbb{N}_0\), there exist \(C,h>0\) (respectively, for every \(h>0\), there exists \(C>0\)) such that \[\begin{align} |(1-\Delta_\xi)^{N_1}(1-\Delta_y)^{N_2}(p(x,y,\xi)u(y))| & \leq 2^{2N_2}Ch^{2N_1+2N_2+N_3} A_{2N_1}B_{2N_2} B_{2N_3}\\ & \times \langle\xi\rangle^{\kappa(2N_2)}\langle (x,y)\rangle^{\lambda(2N_1)}\langle y\rangle^{-2N_3}, \end{align}\] where the factor \(2^{2N_2}\) can be incorporated into \(h\). Also, notice that \[\begin{align} \int_{\mathbb{R}^n}\langle (x,y)\rangle^{\lambda(2N_1)}\langle y\rangle^{-2N_3}\,\mathrm{d}\bar{}y & = \langle x\rangle^{\lambda(2N_1)+n}\int_{\mathbb{R}^n}\langle y\rangle^{\lambda(2N_1)}(1+\langle x\rangle^2|y|^2)^{-N_3}\,\mathrm{d}\bar{}y\\ & \leq C'\langle x\rangle^{\lambda(2N_1)+n}, \end{align}\] where we used the substitution \(y=y'\langle x\rangle\), \(\mathrm{d}\bar{}y = \langle x\rangle^n\mathrm{d}\bar{}y'\), and \[C'=\int_{\mathbb{R}^n}\langle y\rangle^{\lambda(2N_1)-2N_3}\,\mathrm{d}\bar{}y,\] since we can choose \(N_3\) depending on \(N_1\) to be sufficiently large such that the integral converges.

Now, since we have \(\lim_{j\to\infty}\lambda(j)-j=-\infty\), there are just finitely many values of \(j\in\mathbb{N}_0\) such that \(\lambda(j)-j\) is positive. Hence, we can obtain \(\tilde{N}\in\mathbb{N}\) such that \(\tilde{N}>\lambda(j)-j\) for every \(j\in\mathbb{N}_0\). In this case, for each \(N_1\in\mathbb{N}_0\), we can choose \(N_3=N_1+\tilde{N}+n\). Then we have \[\int_{\mathbb{R}^n}\langle y\rangle^{\lambda(2N_1)-2N_3}\,\mathrm{d}\bar{}y = \int_{\mathbb{R}^n}\langle y\rangle^{\lambda(2N_1)-2N_1-2\tilde{N}}\langle y\rangle^{-2n}\,\mathrm{d}\bar{}y,\] where \(\langle y\rangle^{\lambda(2N_1)-2N_1-2\tilde{N}}\leq 1\) for all \(y\in\mathbb{R}^n\) and for all \(N_1\in\mathbb{N}_0\), and the integral \[\int_{\mathbb{R}^n}\langle y\rangle^{-2n}\,\mathrm{d}\bar{}y\] is finite. Hence, we obtain a constant \(C'>0\), independent from \(N_1\), such that \[\int_{\mathbb{R}^n}\langle (x,y)\rangle^{\lambda(2N_1)}\langle y\rangle^{-2N_3}\,\mathrm{d}\bar{}y \leq C'\langle x\rangle^{\lambda(2N_1)+n}\] for all \(N_1\in\mathbb{N}_0\), where \(N_3\) is chosen as before. In this case, 17 is dominated by \[\label{est95I} C_0\langle\xi\rangle^{\kappa(2N_2)-2N_2}\langle x\rangle^{n+\lambda(2N_1)-2N_1},\tag{18}\] where \[C_0 = CC'h^{2N_1+2N_2+2N_3} A_{2N_1}B_{2N_2} B_{2N_3}.\]

Then, for a sufficiently large \(N\), the integral \[\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{i\xi\cdot(x-y)}p(x,y,\xi)u(y)\,\mathrm{d}\bar{}y\,\mathrm{d}\bar{}\xi\] is finite. In particular, the previous estimates allows us to differentiate under the integral sign. Hence, for each \(u\in{\mathcal{S}}_{[{\mathcal{M}}]}({\mathbb{R}}^n)\), \({\mathrm{Op}}(p)u\) defines a function in \({\mathcal{S}}_{[{\mathcal{M}}]}({\mathbb{R}}^n)\).

We now show the continuity of \({\mathrm{Op}}(p)\). Let \(\mathscr{B}\subset{\mathcal{S}}_{[{\mathcal{M}}]}({\mathbb{R}}^n)\) be a bounded subset. Then, there exist \(C,h>0\) (respectively, for every \(h>0\), there exists \(C>0\)) such that \[\sup_{x\in{\mathbb{R}}^n}|x^\beta\partial_x^\alpha u(x)|\leq Ch^{|\alpha|+|\beta|}M_{|\alpha|}M_{|\beta|},\] for all \(\alpha,\beta\in\mathbb{N}_0^n\) and \(u\in\mathscr{B}\). Let us show that \(\mathrm{Op}(p)(\mathscr{B})\) is bounded in \(\mathcal{S}_{[{\mathcal{M}}]}(\mathbb{R}^n)\). Indeed, for every \(\alpha,\beta\in\mathbb{N}_0^n\), we have \[\begin{align} x^\beta D_x^\alpha\mathrm{Op}(p)u(x) & = x^\beta D_x^\alpha\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{i\xi\cdot (x-y)}p(x,y,\xi) u(y)\,\mathrm{d}\bar{}y\,\mathrm{d}\bar{}\xi\\ & = x^\beta\sum_{\gamma\leq\alpha}\binom{\alpha}{\gamma} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} e^{i\xi\cdot (x-y)}\xi^\gamma D_x^{\alpha-\gamma}p(x,y,\xi) u(y)\,\mathrm{d}\bar{}y\,\mathrm{d}\bar{}\xi. \end{align}\]

By the second part of the proof of [3], integration by parts and Leibniz rule give that \(x^\beta D_x^\alpha[\mathrm{Op}(p)u](x)\) is a linear combination of terms of the form \[\label{terms95ST} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} e^{i\xi\cdot(x-y)} (x-y)^{\alpha_1-\gamma_1} \,D^{\gamma_2}_y(u(y)y^{\alpha_2}) \,D_x^{\beta_2}D_y^{\gamma_3}p(x,y,\xi)\,\mathrm{d}\bar{}y\,\mathrm{d}\bar{}\xi,\tag{19}\] where \(\alpha_1+\alpha_2=\alpha\), \(\beta_1+\beta_2=\beta\), and \(\gamma_1+\gamma_2+\gamma_3=\beta_1\). Then, the estimates obtained in the proofs of Propositions 28 and 29, as well as the bound 18 , give us constants \(C,h>0\) such that the terms of the form 19 are bounded by \[Ch^{|\alpha|+|\beta|} M_{|\alpha|}M_{|\beta|}\langle x \rangle^{\lambda(|\beta|)-N},\] for an arbitrary, sufficiently large \(N\in\mathbb{N}_0\). Then, we obtain \[\sup_{x\in{\mathbb{R}}^n}|x^\beta D_x^\alpha\mathrm{Op}(p)u(x)|\leq Ch_0^{|\alpha|+|\beta|}M_{|\alpha|}M_{|\beta|}\langle x\rangle^{n+\lambda(|\beta|)},\] for a new constant \(h_0>0\), that comprises the factor of the form \(h_1^{|\alpha|+|\beta|}\) arising from the sum of the binomial coefficients of Leibniz rule. It follows that \(\mathrm{Op}(p)\) is bounded from \({\mathcal{S}}_{[{\mathcal{M}}]}({\mathbb{R}}^n)\) to itself.

Finally, the proof that \({\mathrm{Op}}(p)\) extends to a continuous operator from \(\mathcal{S}'_{[{\mathcal{M}}]}({\mathbb{R}}^n)\) to itself is analogous to the one of Proposition 1. The proof is complete. ◻

Proof. The claim follows by a straightforward variant of the argument in [4]. ◻

Proof. Since \(p\) is hypoelliptic, let \(q_0\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{-m',-\mu'}({\mathbb{R}}^n)\) be such that \(q_0(x,\xi)=p(x,\xi)^{-1}\) for \((x,\xi)\in Q_B^c\). We claim that \({\mathrm{Op}}(q_0){\mathrm{Op}}(p)=I+{\mathrm{Op}}(r_1)\), where \(r_1\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{-1,-1}({\mathbb{R}}^n)\). The proof is an adaptation of the usual parametrix construction argument. For the sake of completeness, we provide some detail.

By Theorem 26, \({\mathrm{Op}}(q_0){\mathrm{Op}}(p)\) has symbol with asymptotic expansion \[1+\sum_{\alpha\neq 0}\frac{1}{\alpha!}\partial_\xi^\alpha [p(x,\xi)^{-1}] D_x^\alpha p(x,\xi).\]

Notice that, for each \(\alpha\neq 0\), the symbol \(\partial_\xi^\alpha [p(x,\xi)^{-1}] D_x^\alpha p(x,\xi)\) belongs to \({{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{-|\alpha|,-|\alpha|}({\mathbb{R}}^n)\). Indeed, since \(p\) is hypoelliptic, by ?? , Lemma 9, and the Leibniz rule, we obtain \[\begin{align} |\partial_\xi^{\alpha'}\partial_x^\beta(\partial_\xi^\alpha &[p(x,\xi)^{-1}] D_x^\alpha p(x,\xi))|\\ &\leq\; \sum_{\gamma\leq\alpha'}\binom{\alpha'}{\gamma} \sum_{\delta\leq\beta}\binom{\beta}{\delta} |\partial_\xi^{\gamma+\alpha}\partial_x^{\delta} [p(x,\xi)^{-1}]| |\partial_\xi^{\alpha'-\gamma+\alpha}\partial_x^{\beta-\delta+\alpha} p(x,\xi)|\\ &\leq\;\sum_{\gamma\leq\alpha'}\binom{\alpha'}{\gamma} \sum_{\delta\leq\beta}\binom{\beta}{\delta} Ch^{|\alpha|+|\gamma|+|\delta|}\frac{A_{\alpha}A_{\gamma}B_{|\delta|}}{|p(x,\xi)|}\langle\xi\rangle^{-|\alpha+\gamma|}\langle x\rangle^{-|\delta|}\\ & \times h^{|\alpha'-\gamma|+|\beta-\delta|+|\alpha|} A_{|\alpha'-\gamma|}B_{|\beta-\delta|}B_{|\alpha|}|p(x,\xi)|\langle\xi\rangle^{-|\alpha'-\gamma|}\langle x\rangle^{-|\beta-\delta|-|\alpha|}\\ &\leq\;\sum_{\gamma\leq\alpha'}\binom{\alpha'}{\gamma} \sum_{\delta\leq\beta}\binom{\beta}{\delta} Ch^{2|\alpha|+|\alpha'|+|\beta|}A_{|\alpha'|}B_{|\beta|}A_{|\alpha|}B_{|\alpha|}\langle\xi\rangle^{-|\alpha|-|\alpha'|}\langle x\rangle^{-|\alpha|-|\beta|}\\ &\leq\;C_0h_0^{|\alpha'|+|\beta|}A_{|\alpha'|}B_{|\beta|}\langle\xi\rangle^{-|\alpha|-|\alpha'|}\langle x\rangle^{-|\alpha|-|\beta|}, \end{align}\] where \(C_0=Ch^{2|\alpha|}A_{|\alpha|}B_{|\alpha|}\) and \(h_0=2h\), for the inductive case. For the projective case, for each \(h>0\), we can choose \(h/2\) in the symbol estimates.

In this case, by Proposition 19, we take \(r_1\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{-1,-1}({\mathbb{R}}^n)\) such that \[r_1(x,\xi) \sim \sum_{\alpha\neq 0}\frac{1}{\alpha!}\partial_\xi^\alpha [p(x,\xi)^{-1}] D_x^\alpha p(x,\xi),\] which proves our claim. In particular, \({\mathrm{Op}}(q_0){\mathrm{Op}}(p)\) is an operator with symbol in \({{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{0,0}({\mathbb{R}}^n)\).

Now, we set \(r_0(x,\xi)=1\) and, for every \(j\in{\mathbb{N}}\), let \(r_j(x,\xi)\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{-j,-j}({\mathbb{R}}^n)\) be the symbol of \({\mathrm{Op}}(r_1)^j\). Again by Proposition 19, let \(r\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{0,0}({\mathbb{R}}^n)\) be such that \[r(x,\xi) \sim \sum_{j\in{\mathbb{N}}_0}r_j(x,\xi).\]

Then, by a standard argument, if we take \(q\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{-m',-\mu'}({\mathbb{R}}^n)\) to be the symbol of \({\mathrm{Op}}(r){\mathrm{Op}}(q_0)\), we obtain \({\mathrm{Op}}(q){\mathrm{Op}}(p)=I+R_1\), with \(R_1\) \([{\mathcal{M}}]\)-regularizing. Analogously, we obtain \(q'\in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{-m',-\mu'}({\mathbb{R}}^n)\) and \(R_2\) \([{\mathcal{M}}]\)-regularizing such that \({\mathrm{Op}}(p){\mathrm{Op}}(q')=I+R_2\), and \({\mathrm{Op}}(q)-{\mathrm{Op}}(q')\) is \([{\mathcal{M}}]\)-regularizing. The details are left for the reader. ◻

Proof. The proof is identical to that of [8]. Point (a) follows from the mean value theorem, while point (b) follows from the fact that \(\nabla\varphi(x,x)=\varphi'(x)\) and \((I1)\) holds. Finally, for point (c), we choose a function \(g\in\mathcal{D}_{[{\mathcal{B}}]}(\mathbb{R})\) such that \(g(t)=1\) if \(t\leq 1/2\) and \(g(t)=0\) if \(t>1\). Then, the function \[\chi(x,y) = g\left(\frac{|x-y|}{k\langle x\rangle}\right)\] satisfies the desired properties. ◻

Proof. First, by Peetre’s inequality, we have that \[\label{peetre} \langle x-t(y-x)\rangle^{-j} \leq 2^j\langle x\rangle^{-j}\langle x-y\rangle^{j},\tag{21}\] for all \(t\in[0,1]\), \(x,y\in\mathbb{R}^n\), and \(j\geq 0\).

(a) Write \[\varphi' = \begin{bmatrix} \partial_{x_1}\varphi_1 & \cdots & \partial_{x_n}\varphi_1 \\ \vdots & \ddots & \vdots \\ \partial_{x_1}\varphi_n & \cdots & \partial_{x_n}\varphi_n \end{bmatrix} = \begin{bmatrix} \varphi_1^{(1)} & \cdots & \varphi_1^{(n)} \\ \vdots & \ddots & \vdots \\ \varphi_n^{(1)} & \cdots & \varphi_n^{(n)} \end{bmatrix}.\]

Then, we have \[\det\varphi' = \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\varphi_1^{(\sigma(1))}\cdots\varphi_n^{(\sigma(n))},\] where \(S_n\) is the symmetric group of order \(n\), which has \(n!\) elements. By \((I1)\), there exist \(C,h>0\) such that \[|\partial_x^\alpha\varphi_\ell^{(\sigma(\ell))}(x)|\leq Ch^{|\alpha|}\alpha!\langle x\rangle^{-|\alpha|},\] for all \(\alpha\in\mathbb{N}_0^n\), \(x\in\mathbb{R}^n\), \(\ell=1,\dots,n\), and \(\sigma\in S_n\). Let \(\alpha_1,\cdots,\alpha_n\in\mathbb{N}_0^n\) be such that \(\alpha_1+\cdots+\alpha_n=\alpha\). Then, we have \[|\partial_x^{\alpha_1}\varphi_1^{(\sigma(1))}\cdots \partial_x^{\alpha_n}\varphi_n^{(\sigma(n))}| \leq (Ch^{|\alpha_1|}\alpha_1!\langle x\rangle^{-|\alpha_1|})\cdots (Ch^{|\alpha_n|}\alpha_n!\langle x\rangle^{-|\alpha_n|}) \leq C^nh^{|\alpha|}\alpha!\langle x\rangle^{-|\alpha|}.\]

Given \(j,k\in\{1,\dots,n\}\), we also see that \[\partial_{x_j}\det\varphi'(x) = \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\sum_{\ell=1}^n \varphi_1^{(\sigma(1))}\cdots \partial_{x_j}\varphi_{\ell}^{(\sigma(\ell))}\cdots \varphi_n^{(\sigma(n))},\] and \[\partial_{x_k}\partial_{x_j}\det\varphi'(x) = \sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\sum_{\nu=1}^n \sum_{\ell=1}^n \varphi_1^{(\sigma(1))}\cdots \partial_{x_k}\varphi_{\nu}^{(\sigma(\nu))} \cdots \partial_{x_j}\varphi_{\ell}^{(\sigma(\ell))}\cdots \varphi_n^{(\sigma(n))}.\]

By induction, it follows that \(\partial_x^\alpha\det\varphi'(x)\) is a linear combination of \(n!n^{|\alpha|}\) terms of the form \[\mathrm{sgn}(\sigma)\partial_x^{\alpha_1}\varphi_1^{(\sigma(1))}\cdots \partial_x^{\alpha_n}\varphi_n^{(\sigma(n))},\] with \(\alpha_1+\cdots+\alpha_n=\alpha\). Hence, we conclude \[|\partial_x^\alpha\det\varphi'(x)|\leq C_0h_0^{|\alpha|}\alpha!\langle x\rangle^{-|\alpha|},\] for all \(\alpha\in\mathbb{N}_0^n\) and \(x\in\mathbb{R}^n\), where \(C_0=C^nn!\) and \(h_0=nh\). From this estimate and 21 , we obtain \(C,h>0\) such that \[|\partial_x^\beta\partial_y^\gamma G_t(x,y)|\leq Ch^{|\beta|+|\gamma|}(\beta!\gamma!)\langle x\rangle^{-|\beta|-|\gamma|}\langle x-y\rangle^{|\beta|+|\gamma|},\] for all \(\beta,\gamma\in\mathbb{N}_0^n\), \(x,y\in\mathbb{R}^n\), and \(t\in [0,1]\). Notice that the absolute value in the definition of \(G_t(x,y)\) does not affect the smoothness of the map since \(\varphi\) is a diffeomorphism. Finally, we apply Proposition 2 to obtain the desired estimate.

(b) Using the notation \(\varphi_{\ell}^{(k)}=\partial_{x_k}\varphi_\ell\) from the previous point, by \((I1)\) and 21 it follows \[\begin{align} \left|\partial_x^\beta\int_0^1\varphi_{\ell}^{(\sigma(\ell))}(x+t(y-x))\,\mathrm{d}t\right| & \leq \int_0^1\left|\partial_x^\beta \varphi_{\ell}^{(\sigma(\ell))}(x+t(y-x))\right|\,\mathrm{d}t \\ & \leq C(2h)^{|\beta|}\beta!\langle x\rangle^{-|\beta|}\langle x-y\rangle^{|\beta|}. \end{align}\]

For the \(y\)-derivatives, we obtain analogous estimates. Then, the estimates for \(H_t(x,y)\) follow from those obtained in (a) and Proposition 2.

(c) Fix \(t\in [0,1]\) and write \(X=(X_1,\dots,X_n)\), \(Y=(Y_1,\dots,Y_n)\), \(\Xi=(\Xi_1,\dots,\Xi_n)\), where \[X_j=\varphi_j(x),\quad Y_j=\varphi_j(x+t(y-x)),\quad\text{and}\quad \Xi_j=\sum_{k=1}^n \nabla\varphi(x,y)_{jk}^{-T}\xi_j,\quad j=1,\dots,n.\]

In the sequel, we write \[p_1^\sharp(x,y,\xi)=p_1(\varphi(x),\varphi(x+t(y-x)),\nabla\varphi(x,y)^{-T}\xi)=p_1(X,Y,\Xi).\]

By the chain rule, \[\label{dxp} \begin{align} \partial_{x_j}p_1^\sharp &= \sum_{k=1}^n \frac{\partial p_1}{\partial X_k}\frac{\partial X_k}{\partial x_j} + \sum_{k=1}^{n}\frac{\partial p_1}{\partial Y_k}\frac{\partial Y_k}{\partial x_j} + \sum_{k=1}^{n}\frac{\partial p_1}{\partial \Xi_k}\frac{\partial \Xi_k}{\partial x_j}, \\ \partial_{y_j}p_1^\sharp &= \sum_{k=1}^{n}\frac{\partial p_1}{\partial Y_k}\frac{\partial Y_k}{\partial y_j} + \sum_{k=1}^{n}\frac{\partial p_1}{\partial \Xi_k}\frac{\partial \Xi_k}{\partial y_j}, \\ \partial_{\xi_j}p_1^\sharp &= \sum_{k=1}^n \frac{\partial p_1}{\partial\Xi_k}\nabla\varphi(x,y)_{kj}^{-T}. \end{align}\tag{22}\]

Observe that, by \((I1)\), it holds \[\langle\varphi(x)\rangle \asymp \langle x\rangle\quad\text{and}\quad \langle \nabla\varphi(x,y)^{-T}\xi\rangle \asymp \langle\xi\rangle.\]

Then, the fact that \(p_1\in {\mathrm{AG}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m,0,\mu}(\mathbb{R}^n)\) and 21 , together with \((I1)\), imply \[\left|\frac{\partial p_1}{\partial X_j}\right| \leq Ch B_1\langle x\rangle^{m-1} \langle \xi\rangle^{\mu},\quad \left|\frac{\partial p_1}{\partial Y_j}\right| \leq Ch B_1 \langle x\rangle^{m-1} \langle x-y\rangle \langle \xi\rangle^{\mu},\] and \[\left|\frac{\partial p_1}{\partial \Xi_j}\right| \leq Ch A_1 \langle x\rangle^{m} \langle\xi\rangle^{\mu-1}.\]

We also have, again by \((I1)\), \[\left|\frac{\partial X_j}{\partial x_j}\right| \leq Ch B_1 ,\quad \left|\frac{\partial Y_j}{\partial x_j}\right| \leq Ch B_1,\] \[\left| \frac{\nabla\varphi(x,y)_{k\ell}^{-T}}{\partial x_j}\right| \leq Ch B_1\langle x\rangle^{-1}\langle x-y\rangle, \quad \left|\frac{\partial Y_j}{\partial y_j}\right| \leq Ch B_1\] \[\left|\frac{\nabla\varphi(x,y)_{k\ell}^{-T}}{\partial y_j}\right| \leq Ch B_1\langle x\rangle^{-1}\langle x-y\rangle,\quad |\nabla\varphi(x,y)_{kj}^{-T}|\leq C.\]

Notice that the terms \(\xi_\ell\) that arise when we differentiate \(p_1^\sharp\) with respect to \(x_j\) or \(y_j\) are compensated by the decay given by \(\partial p_1^\sharp/\partial\Xi_\ell\). Then, combining the previous estimates, we have proven \[\label{est951st} \left| \partial^\alpha_x\partial^\beta_y\partial^\gamma_\xi p_1^\sharp(x,y,\xi) \right|\leq (C^23nh)h^1 A_{|\alpha|}B_{|\beta|}B_{|\gamma|} \langle x\rangle^{m-|\alpha|-|\beta|}\langle x-y\rangle^{|\alpha|+|\beta|} \langle\xi\rangle^{\mu-|\gamma|},\tag{23}\] for \(|\alpha|+|\beta|+|\gamma|\leq 1\).

If we differentiate 22 with respect to \(x_\ell\), the Leibniz rule yields \[\begin{align} \label{dxxp} \partial_{x_\ell}\partial_{x_j}p_1^\sharp = & \sum_{k=1}^n \left( \left(\partial_{x_\ell}\frac{\partial p_1}{\partial X_k}\right)\frac{\partial X_k}{\partial x_j}+\frac{\partial p_1}{\partial X_k}\frac{\partial^2 X_k}{\partial x_\ell\partial x_j}\right) + \sum_{k=1}^{n}\left( \left(\partial_{x_\ell}\frac{\partial p_1}{\partial Y_k}\right)\frac{\partial Y_k}{\partial x_j}+\frac{\partial p_1}{\partial Y_k}\frac{\partial^2 Y_k}{\partial x_\ell\partial x_j}\right)\\ + &\sum_{k=1}^{n}\left( \left(\partial_{x_\ell}\frac{\partial p_1}{\partial \Xi_k}\right)\frac{\partial \Xi_k}{\partial x_j}+\frac{\partial p_1}{\partial \Xi_k}\frac{\partial^2 \Xi_k}{\partial x_\ell\partial x_j}\right). \end{align}\tag{24}\]

Since \(p_1\in {\mathrm{AG}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m,0,\mu}\), it follows \[\frac{\partial p_1}{\partial X_k}(X,Y,\Xi)\in {\mathrm{AG}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m-1,0,\mu},\; \frac{\partial p_1}{\partial Y_k}(X,Y,\Xi)\in {\mathrm{AG}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m,-1,\mu},\; \frac{\partial p_1}{\partial \Xi_k}(X,Y,\Xi)\in {\mathrm{AG}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m,0,\mu-1}.\]

Moreover, these symbols satisfy \[\left|\partial^\alpha_X\partial^\beta_Y\partial^\gamma_\Xi \frac{\partial p_1}{\partial X_k}\right| \leq Ch^{|\alpha|+|\beta|+|\gamma|+1} A_{|\alpha|}B_{|\beta|+1}B_{|\gamma|} \langle x\rangle^{m-|\alpha|-1}\langle y\rangle^{-|\beta|} \langle \xi\rangle^{\mu-|\gamma|},\] \[\left|\partial^\alpha_X\partial^\beta_Y\partial^\gamma_\Xi \frac{\partial p_1}{\partial Y_k}\right| \leq Ch^{|\alpha|+|\beta|+|\gamma|+1}A_{|\alpha|} B_{|\beta|}B_{|\gamma|+1} \langle x\rangle^{m-|\alpha|}\langle y\rangle^{-|\beta|-1}\langle \xi\rangle^{\mu-|\gamma|},\] \[\left| \partial^\alpha_X\partial^\beta_Y\partial^\gamma_\Xi \frac{\partial p_1}{\partial \Xi_k}\right|\leq Ch^{|\alpha|+|\beta|+|\gamma|+1}A_{|\alpha|+1} B_{|\beta|}B_{|\gamma|} \langle x\rangle^{m-|\alpha|}\langle y\rangle^{-|\beta|}\langle \xi\rangle^{\mu-|\gamma|-1},\] for \(\alpha,\beta,\gamma\in\mathbb{N}_0^n\), where the constants \(C,h>0\) are the same appearing in the previous estimates above. If we now apply 23 to those symbols in place of \(p_1^\sharp\), we obtain \[\begin{align} \left| \partial_{x_\ell}\frac{\partial p_1}{\partial X_k} \right| &\leq (C^23nh)h^2 B_2 \langle x\rangle^{m-2}\langle \xi\rangle^{\mu}, \\ \left| \partial_{x_\ell}\frac{\partial p_1}{\partial Y_k} \right| &\leq (C^23nh)h^2 B_2\langle x\rangle^{m-2}\langle x-y\rangle^2 \langle \xi\rangle^{\mu}, \\ \left| \partial_{x_\ell}\frac{\partial p_1}{\partial \Xi_k} \right| &\leq (C^23nh)h^2 A_1 B_1 \langle x\rangle^{m-1}\langle x-y\rangle^2. \end{align}\]

By \((I1)\) for the derivatives of \(X_j\), \(Y_j\), and \(\Xi_j\), \(\partial_{x_\ell}\partial_{x_j}p_1^\sharp\) is a sum of \(2(3n)\) terms satisfying the desired estimates. By an induction argument, a derivative of order \(N\) is a sum of \(2^{N-1}3n\) terms. Then, incorporating the factor \(1/2\) within the constant \(C_0=C^23nh\) and \(2^N\) within the constant \(h\), we obtain the desired estimate for the \(x\)-derivatives. Similar arguments give the estimates for the derivatives with respect to \(y\) and \(\xi\). This proves the estimates in the inductive case.

The estimates for the projective case can be proved in a similar fashion, by choosing the constants \(h>0\) in an appropriate way, to absorb the extra constants that appear during the computations. The proof is complete. ◻

Proof. Write \(\mathrm{Op}(q)-\mathrm{Op}(\tilde{q}) = \mathrm{Op}(q-\tilde{q}_N) - \mathrm{Op}(\tilde{q}-\tilde{q}_N)\). By definition, we have \(\tilde{q}-\tilde{q}_N \in {{\mathrm{SG}}}_{[{\mathcal{A}},{\mathcal{B}}]}^{m-N,\mu-N}\), hence \(\tilde{q}-\tilde{q}_N\) satisfies ?? . Moreover, for every \(u\in{\mathcal{S}}_{[{\mathcal{M}}]}(\mathbb{R}^n)\), we have \[\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} e^{i\xi\cdot(x-y)}(q(x,y,\xi)-\tilde{q}_N(x,\xi))u(y)\,\mathrm{d}y\, \mathrm{d}\xi = \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} e^{i\xi\cdot(x-y)}q_N(x,y,\xi)u(y)\,\mathrm{d}y\, \mathrm{d}\xi,\] by 25 and the definition of \(q_N\). Finally, 26 gives us that \(q-\tilde{q}_N\) also satisfies ?? . ◻

Proof. If we choose \(N\in\mathbb{N}\) sufficiently large, the integral in ?? exists for every \(x,y\in\mathbb{R}^n\). Again, since we can choose \(N\) arbitrarily large, we can differentiate \(K\) under the integral sign. By ?? , we find \(K\in{\mathcal{S}}_{[{\mathcal{M}}]}(\mathbb{R}^{2n})\), which implies that \(\mathrm{Op}(r)\) is \([{\mathcal{M}}]\)-regularizing. ◻

Proof. Point (a) is consequence of the regularizing property of the convolution, while Points (b) and (c) are straightforward. Concerning Point (d), since \(\Phi_\varepsilon\) is compactly supported, we have \[\begin{align} |\partial_x^\beta\Phi_U(x)| & \leq \int_{\mathbb{R}^n}\chi_U(y)|\partial_x^\alpha \Phi_\varepsilon(x-y)|\,\mathrm{d}y\\ & \leq \sup_{y\in\mathbb{R}^n}|\partial_x^\alpha\Phi_\varepsilon(x-y)|\int_{\text{supp}(\Phi_\varepsilon(x-y))}\,\mathrm{d}y\\ & \leq C_\varepsilon\sup_{x\in\mathbb{R}^n}|\partial_x^\alpha\Phi_\varepsilon(x)|, \end{align}\] for all \(\beta\in{\mathbb{N}}_0^n\), where \(C_\varepsilon>0\) is the volume of the support of \(\Phi_\varepsilon\). Hence, it is enough to show that \(\Phi_\varepsilon\) satisfies ?? . Recall that \(\Phi_\varepsilon\) decays exponentially. Then, for every \(\beta\in\mathbb{N}_0^n\), we have that \(|\partial_x^{\beta}\Phi_\varepsilon|\leq \langle x\rangle^{-|\beta|}\) outside a fixed compact set independent of \(\beta\), while inside such compact set we have \[|\partial_x^{\beta}\Phi_\varepsilon|\leq Ch^{|\beta|}B_{|\beta|},\] since \(\Phi_\varepsilon\in\mathcal{E}_{[{\mathcal{B}}]}({\mathbb{R}}^n)\). Combining both estimates, we obtain ?? . ◻

Proof. For each \(k=1,\dots,N\), define on \(U_k^\sharp\) the function \(\Psi_k(x)=\Phi_{(U_k)_\varepsilon}(\varepsilon,x)\), where \(\varepsilon=\delta/3\). Here, \(\delta>0\) is the constant from condition \((SGA2)\). Then, using the local charts, set \[\Psi_k^*(x) = \begin{cases} \Psi_k(\varphi_k(x)),& x\in\Omega_k^\sharp, \\ 0,& x\notin\Omega_k^\sharp. \end{cases}\] and, for each \(k=1,\dots,N\), define \[\Phi_k:X\to \mathbb{R}\colon x\mapsto \Phi_k(x) = \frac{\Psi_k^*(x)}{\sum_{\ell=1}^N\Psi_\ell^*(x)}.\]

Notice that each \(\Phi_k\) is well-defined, since the denominator is always bigger than \(1\). Moreover, it is clear, by the definition of \(\Phi_k\), that \(\sum_{k=1}^N\Phi_k=1\). Moreover, \(0\leq \Phi_k\leq 1\), by Corollary 3, (b). Then, \(\{\Phi_k\}_{k=1}^N\) is a partition of unity subordinated to \(\{\Omega_k^\sharp\}_{k=1}^N\). Finally, ?? is proved employing condition \((I1)\) and Corollary 3, (a). ◻

Proof. The proof is analogous to that of [8]. ◻

Proof. In an interior chart \((\Omega_k,\varphi_k)\), we just set \(\varrho_k\equiv 1\). In a boundary chart \((\Omega_k,\varphi_k)\), with \(\varphi_k(\Omega_k)\subset\mathbb{R}^{n-1}\times[0,+\infty)\), we consider \(\varrho_k(x_1,\dots,x_n)=x_n\). Then, \(\varrho_k=0\) in \(\partial X\) and \(\varrho_k>0\) outside \(\partial X\). Let \(\{\Phi_k\}_{k=1}^N\) be a partition of unity of order subordinated to the cover \(\{\Omega_k\}_{k=1}^N\), such that each \(\Phi_k\) belongs to \({\mathcal{D}}_{[{\mathcal{M}}]}({\overline{X}})\). Since each \(\varrho_k\) is analytic on its domain, \[\varrho(x) = \sum_{k=1}^N \Phi_k(x)(\varrho_k\circ\varphi_k^{-1})(x)\] belongs to \({\mathcal{E}}_{[{\mathcal{M}}]}({\overline{X}})\). Moreover, \(\mathrm{d}\varrho\) does not vanish in any \(y\in \partial X\) as shown in the proof of [41]. ◻

Proof. We just need to check the behaviour of the functions in the exit charts, where we have coordinates of the form \((t,ty)\), where \(t>\delta\) and \(y \in \tilde{U}_k\subset{\mathbb{R}}^{n-1}\), for \(k=1,\dots,N\) and \(\delta\geq 1\). Here, \(t=1/\varrho\) and \(\tilde{U}_k\) is a local chart of the closed manifold \(\partial X_0\). By the compactness of \(\partial X_0\), we can assume that each \(\tilde{U}_k\) is bounded. In this case, we have \[|(t,ty)| = (t^2+t^2y^2)^{\frac{1}{2}} = t\langle y\rangle \asymp t,\] that is, we can obtain a constant \(\kappa>0\) such that \[\label{ty95asymp95t} \kappa^{-1}|(t,ty)| \leq t \leq \kappa|(t,ty)|.\tag{29}\]

First, suppose that \(f\in\mathcal{S}_{\{{\mathcal{M}}\}}(X)\) and let \(h,\varepsilon>0\) be such that \[\sup_{(t,y)\in U_k}\sup_{\alpha\in{\mathbb{N}}_0^n}\frac{e^{\omega_{\mathcal{M}}(\varepsilon|(t,ty)|)}|\partial^\alpha f(t,y)|}{h^{|\alpha|}M_{|\alpha|}} < +\infty.\]

Write \(f(x) = e^{-\omega_{\mathcal{M}}(\varepsilon/(4H\kappa\varrho))} e^{\omega_{\mathcal{M}}(\varepsilon/(4H\kappa\varrho))} f(x)\), where \(H\geq 1\) is given by \((M2)'\) and \(\kappa>0\) is given by 29 .

By Propositions, 3, 5, and 15, we have \(e^{\omega_{\mathcal{M}}(\varepsilon/(4H\kappa\varrho))} f(x)=g(x)\in{\mathcal{E}}_{\{{\mathcal{M}}\}}(X)\). In any exit chart, by Leibniz Rule and Faà di Bruno Formula (Lemma 4), we obtain \[\begin{align} |\partial_t^\alpha\partial_y^\beta g(t,y)| & \leq \sum_{k=0}^\alpha\binom{\alpha}{k}|\partial_t^{\alpha-k}\partial_y^\beta f(t,y)||\partial_t^k e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H\kappa}t\right)}|\\ & \leq \sum_{k=0}^\alpha\binom{\alpha}{k}e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H\kappa}t\right)}|\partial_t^{\alpha-k}\partial_y^\beta f(t,y)| \sum_{\gamma\in\Delta(k)}\frac{k!}{\gamma!}\prod_{\ell=1}^k\left|\frac{\partial_t^\ell \omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H\kappa}t\right)}{\ell!}\right|^{\gamma_\ell}, \end{align}\] for all \(\alpha\in{\mathbb{N}}_0\) and \(\beta\in{\mathbb{N}}_0^{n-1}\), where the set \(\Delta(k)\) is defined in 5 .

Notice that, since \({\mathcal{M}}\) satisfies \((M4)\), the sequence \(\{M_j/j!\}_{j\in{\mathbb{N}}_0}\) satisfies \((M1)\). Then, by Proposition 15, Lemmas 5 and 6, and the fact that \(t^k\leq k!e^t\), for all \(t>0\) and \(k\in{\mathbb{N}}_0\), we obtain \(C,h>0\) such that \[\begin{align} \sum_{\gamma\in\Delta(k)}\frac{k!}{\gamma!}\prod_{\ell=1}^k\left|\frac{\partial_t^\ell \omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H\kappa}t\right)}{\ell!}\right|^{\gamma_\ell} & \leq \sum_{\gamma\in\Delta(k)}\frac{k!}{\gamma!}\prod_{\ell=1}^k\left|\frac{C\varepsilon^\ell (4H\kappa)^{-\ell}h^\ell M_\ell \omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H}t\right)}{\ell!}\right|^{\gamma_\ell}\\ & \leq h_0^k \sum_{\gamma\in\Delta(k)} \frac{k!}{\gamma!} C^{|\gamma|} \omega_{\mathcal{M}}\left(\tfrac{\varepsilon }{4H\kappa}t\right)^{|\gamma|}\frac{M_k}{k!}\frac{|\gamma|!}{M_{|\gamma|}}\\ & \leq h_0^k \sum_{\gamma\in\Delta(k)} \frac{1}{\gamma!} C^{|\gamma|}|\gamma|! e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon }{4H\kappa}t\right)} M_k\frac{|\gamma|!}{M_{|\gamma|}}\\ & \leq C_{\mathcal{M}}h_0^k M_k e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H\kappa}t\right)} \sum_{\gamma\in\Delta(k)} \frac{|\gamma|!}{\gamma!} C^{|\gamma|}\\ & \leq C_{\mathcal{M}}h_0^k M_k e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H\kappa}t\right)} C(1+C)^{k-1}\\ & \leq C_{\mathcal{M}}(h_0C_0)^kM_k e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon }{4H\kappa}t\right)}, \end{align}\] where \(h_0=h\varepsilon(4H\kappa)^{-1}\) and \(C_0=1+C\). Here, the constant \(C_{\mathcal{M}}>0\) is such that \(|\gamma|!/M_{|\gamma|}\leq C_{\mathcal{M}}\), for all \(\gamma\), which exists by Proposition 2.

Hence, combining the previous estimates and using Proposition 15, we obtain \[\begin{align} |\partial_t^\alpha\partial_y^\beta g(t,y)| & \leq C_{\mathcal{M}}\sum_{k=0}^\alpha\binom{\alpha}{k} e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H}t\right)} e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{4H}t\right)}|\partial_t^{\alpha-k}\partial_y^\beta f(t,y)|(C_0h_0)^k M_k\\ & \leq C_{\mathcal{M}}\sum_{k=0}^\alpha\binom{\alpha}{k} e^{\omega_{\mathcal{M}}(\varepsilon t)}|\partial_t^{\alpha-k}\partial_y^\beta f(t,y)|(C_0h_0)^k M_k\\ & \leq C_{\mathcal{M}}\sum_{k=0}^\alpha\binom{\alpha}{k} e^{\omega_{\mathcal{M}}(\varepsilon (t+|y|))}|\partial_t^{\alpha-k}\partial_y^\beta f(t,y)|(C_0h_0)^k M_k\\ & \leq C_{\mathcal{M}}\sum_{k=0}^\alpha\binom{\alpha}{k} C_fh_f^{\alpha-k+|\beta|}M_{\alpha-k+|\beta|}(C_0h_0)^kM_k\\ & \leq C_1h_1^{\alpha-k+|\beta|}M_{\alpha+|\beta|}, \end{align}\] for all \(\alpha\in{\mathbb{N}}_0\) and \(\beta\in{\mathbb{N}}_0^{n-1}\), where the constants \(C_f,h_f>0\) are given by the fact that \(f\in\mathcal{S}_{\{{\mathcal{M}}\}}(X)\), \(C_0=C_{\mathcal{M}}C_f\), and \(h_1=2\max\{1,h_f,C_0h_0\}\). Hence \(g\in{\mathcal{E}}_{\{{\mathcal{M}}\}}({\overline{X}})\) and we have \(\mathcal{S}_{\{{\mathcal{M}}\}}(X) \subset \dot{{\mathcal{E}}}_{\{{\mathcal{M}}\}}(X)\).

On the other hand, suppose that \(f\in \dot{{\mathcal{E}}}_{\{{\mathcal{M}}\}}(X)\), that is, \(f=e^{-\omega_{\mathcal{M}}(\varepsilon/\varrho)}g\), for some \(g\in {\mathcal{E}}_{\{{\mathcal{M}}\}}({\overline{X}})\) and \(\varepsilon>0\). By Leibniz Rule, Faà di Bruno Formula, Proposition 15, and the estimate 29 , we have \[\begin{align} e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H\kappa}|(t,ty)|\right)}|\partial_t^\alpha\partial_y^\beta f(t,y)| \leq\;& e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)} \sum_{k=0}^\alpha\binom{\alpha}{k} |\partial_t^{\alpha-k}\partial_y^\beta g(t,y)|e^{-\omega_{\mathcal{M}}(\varepsilon t)}\\ & \times \sum_{\gamma\in\Delta(k)}\frac{k!}{\gamma!}\prod_{\ell=1}^k\left|\frac{\partial_t^\ell \omega_{\mathcal{M}}(\varepsilon t)}{\ell!}\right|^{\gamma_\ell}\\ \leq\;& e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H}t\right)} \sum_{k=0}^\alpha\binom{\alpha}{k} C_gh_g^{\alpha-k+|\beta|}M_{\alpha-k+|\beta|} \\ & \times e^{-\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)}e^{-\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)} \sum_{\gamma\in\Delta(k)}\frac{k!}{\gamma!}\prod_{\ell=1}^k\left|\frac{\partial_t^\ell \omega_{\mathcal{M}}(\varepsilon t)}{\ell!}\right|^{\gamma_\ell}. \end{align}\]

Similarly to the computations of the first inclusion, we obtain \[\begin{align} \sum_{\gamma\in\Delta(k)}\frac{k!}{\gamma!}\prod_{\ell=1}^k\left|\frac{\partial_t^\ell \omega_{\mathcal{M}}(\varepsilon t)}{\ell!}\right|^{\gamma_\ell} & \leq \sum_{\gamma\in\Delta(k)}\frac{k!}{\gamma!}\prod_{\ell=1}^k\left|\frac{C\varepsilon^\ell h^\ell M_\ell \omega_{\mathcal{M}}(\varepsilon t)}{\ell!}\right|^{\gamma_\ell}\\ & \leq (h\varepsilon)^k \sum_{\gamma\in\Delta(k)} \frac{k!}{\gamma!} C^{|\gamma|}\omega_{\mathcal{M}}(\varepsilon t)^{|\gamma|}\frac{M_k}{k!}\frac{|\gamma|!}{M_{|\gamma|}}\\ & = (h \varepsilon)^k \sum_{\gamma\in\Delta(k)} \frac{k!}{\gamma!} C^{|\gamma|}\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H}2H t\right)^{|\gamma|}\frac{M_k}{k!}\frac{|\gamma|!}{M_{|\gamma|}}\\ & \leq (h \varepsilon )^k\sum_{\gamma\in\Delta(k)} \frac{k!}{\gamma!} C^{|\gamma|}K_{\mathcal{M}}\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)^{|\gamma|}\frac{M_k}{k!}\frac{|\gamma|!}{M_{|\gamma|}}\\ & \leq K_{\mathcal{M}}(h\varepsilon)^k \sum_{\gamma\in\Delta(k)} \frac{1}{\gamma!} C^{|\gamma|}|\gamma|!e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)} M_k\frac{|\gamma|!}{M_{|\gamma|}}\\ & \leq K_{\mathcal{M}}C_{\mathcal{M}}(h\varepsilon)^k M_k e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)} \sum_{\gamma\in\Delta(k)} \frac{|\gamma|!}{\gamma!} C^{|\gamma|}\\ & \leq K_{\mathcal{M}}C_{\mathcal{M}}h^k M_k e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)}C(1+C)^{k-1}\\ & \leq K_{\mathcal{M}}C_{\mathcal{M}}(h\varepsilon C_0)^k M_k e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)}, \end{align}\] where \(C_0=1+C\), \(K_{\mathcal{M}}>0\) is such that \(\omega_{\mathcal{M}}(2Ht)\leq K_{\mathcal{M}}\omega_{\mathcal{M}}(t)\) for all \(t\geq 0\), which exists by condition \((W1)\), and \(C_{\mathcal{M}}>0\) is given by Proposition 2.

Combining the the estimates above, we obtain \[\begin{align} e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H\kappa}|(t,ty)|\right)}|\partial_t^\alpha\partial_y^\beta f(t,y)| \leq\;& e^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H}t\right)} \sum_{k=0}^\alpha\binom{\alpha}{k} C_gh_g^{\alpha-k+|\beta|}M_{\alpha-k+|\beta|} \\ & \times e^{-\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)}e^{-\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)}K_{\mathcal{M}}C_{\mathcal{M}}(hC_0)^kM_ke^{\omega_{\mathcal{M}}\left(\tfrac{\varepsilon}{2H} t\right)}\\ \leq\;& C_0 h_0^{\alpha+|\beta|} M_{\alpha+|\beta|}, \end{align}\] where \(C_0=K_{\mathcal{M}}C_{\mathcal{M}}C_g\), \(h_0 = 2\max\{1,h\varepsilon C_0,h_g\}\), and the constants \(C_g,h_g>0\) are given by the fact that \(g\in{\mathcal{E}}_{\{{\mathcal{M}}\}}(\overline{X})\). Hence \(f\in\mathcal{S}_{\{{\mathcal{M}}\}}(X)\) and \(\dot{{\mathcal{E}}}_{\{{\mathcal{M}}\}}(X) \subset\mathcal{S}_{\{{\mathcal{M}}\}}(X)\). The projective case can be proved by similar arguments. ◻

Proof. The proof follows the same lines of the one of [46], employing Proposition 25. ◻

Proof. By Theorem 33, on each local chart we can obtain a local parametrix, from which we can obtain a global parametrix by using the partition of unity and the cutoff functions that we constructed in Theorem 41 and Corollary 4, respectively. This implies ?? . ◻