I am totally new to pseudo-differential operators and I’m wondering if a differential operator is a pseudo-differential operator. So, I want to show , using the definition of the symbol given by Hörmander definition of symbols (see Pic), that if $p(x,\xi) =\sum_{|\alpha|\le m} a_{\alpha}(x) \xi^{\alpha}$ is a symbol of differential operator, it’s also a symbol in $S^m$ but I could not bounded the coefficients in x...
Thanks
Hörmander's definition of a pseudo-differential operator on an open subset $\Omega$ of $\mathbb R^n$ in the class $\text{Op}S^m$ ($m\in \mathbb R$) is the following: take a symbol $a$, that is a smooth function on $\Omega\times\mathbb R^n$ such that \begin{multline} \forall K \text{ compact }\subset \Omega, \forall \alpha, \beta\in \mathbb N^n,\ \exists C_{K\alpha\beta},\ \forall (x,\xi)\in K\times\mathbb R^n,\\ \quad \vert(\partial_x^\alpha\partial_\xi^\beta a)(x,\xi)\vert\le C_{K\alpha\beta} (1+\vert \xi\vert)^{m-\vert \beta\vert}, \tag{1}\end{multline} and consider the operator $\text{Op}(a)$ defined on $C_c^\infty(\Omega)$ by the formula $$ (\text{Op}(a) u)(x)=\int e^{2iπ x\cdot \xi} a(x,\xi)\hat u(\xi) d\xi. $$ Then if you consider a differential operator $P=\sum_{\vert \gamma\vert\le m} p_\gamma(x) D_x^\gamma $ with $p_\gamma\in C^\infty(\Omega)$, you find that it is a pseudo-differential operator of order $m$ on $\Omega$ since if $u\in C^\infty_c(\Omega)$, you have $$ (P u)(x)=\sum_{\vert \gamma\vert\le m}p_\gamma(x)\int e^{2iπ x\cdot \xi}\xi^\gamma \hat u(\xi) d\xi =\int e^{2iπ x\cdot \xi}p(x,\xi) \hat u(\xi) d\xi, $$ with $p(x,\xi)=\sum_{\vert \gamma\vert\le m}p_\gamma(x)\xi^\gamma$; the symbol $p$ obviously verifies (1).
