Symbol class

Remarks

Example: let $a(x, \xi) = \sum_{\abs{\gamma} \le k} a_\gamma(x) \xi^\gamma$, where each $a_\gamma$ is smooth and bounded with all its derivatives. Then $a \in S^k$. Indeed $\partial_\xi^\alpha \xi^\gamma$ is zero or a monomial of degree $\abs{\gamma} - \abs{\alpha}$, so it is bounded by a constant times $(1 + \abs{\xi})^{k - \abs{\alpha}}$. Derivatives in $x$ only change the bounded coefficients. Also $(1 + \abs{\xi}^2)^{s/2} \in S^s$ for every $s \in \R$.

Non-example: $a(x, \xi) = \abs{\xi}$ is not in any symbol class, because it is not smooth at $\xi = 0$. Similarly $a(\xi) = e^{i \abs{\xi}^2}$ is smooth but lies in no class $S^m_{1,0}$, because its derivatives of order $k$ in $\xi$ grow like $\abs{\xi}^k$.

The classes are nested: $S^m_{\rho, \delta} \subseteq S^{m'}_{\rho, \delta}$ for $m \le m'$. Products satisfy $S^m_{\rho, \delta} \cdot S^{m'}_{\rho, \delta} \subseteq S^{m + m'}_{\rho, \delta}$, and $\partial_\xi^\alpha$ maps $S^m_{\rho, \delta}$ into $S^{m - \rho\abs{\alpha}}_{\rho, \delta}$.