Pseudo-differential operator

Remarks

Example: let $P = \sum_{\abs{\alpha} \le k} a_\alpha(x) D^\alpha$ with $D^\alpha = (-i)^{\abs{\alpha}} \partial^\alpha$ and coefficients $a_\alpha$ smooth and bounded with all their derivatives. By Fourier inversion, $D^\alpha u(x) = (2\pi)^{-n} \int e^{i x \cdot \xi} \xi^\alpha \hat{u}(\xi) \, d\xi$, so $P = \mathrm{Op}(p)$ with $p(x, \xi) = \sum_\alpha a_\alpha(x) \xi^\alpha \in S^k$.

Non-example: the operator $u \mapsto \mathcal{F}^{-1}\big( e^{i\abs{\xi}^2} \hat{u} \big)$, a Schrödinger evolution operator, is not a pseudo-differential operator of any order. The symbol of an operator of this form is determined by the operator, here $e^{i \abs{\xi}^2}$, and it is in no class $S^m$.

The composition of $\mathrm{Op}(a)$ with $a \in S^m$ and $\mathrm{Op}(b)$ with $b \in S^{m'}$ is $\mathrm{Op}(c)$ with $c \in S^{m + m'}$, and $c - ab \in S^{m + m' - 1}$. Operators of order $0$ are bounded on $L^2(\R^n)$, and operators of order $-\infty$ are smoothing. The same formula with a symbol in $S^m_{\rho, \delta}$ defines the larger classes $\Psi^m_{\rho, \delta}$.