---
tag: Z9WE435T9G
kind: definition
name: Symbol class
msc: [35S05, 47G30]
defines: AM4RNWVV4Q
uses: [0V33BNKZ9S, 09EGC6F8PT]
sources:
  - url: https://en.wikipedia.org/wiki/Differential_operator
  - url: https://en.wikipedia.org/wiki/Pseudo-differential_operator
---

## Definition

Let $n \ge 1$, let $m \in \R$, and let $0 \le \delta \le \rho \le 1$ with $\delta < 1$.
The *symbol class* $S^m_{\rho, \delta}(\R^n \times \R^n)$ is the set of [[smooth functions|0V33BNKZ9S]] $a \colon \R^n \times \R^n \to \C$, $(x, \xi) \mapsto a(x, \xi)$, such that for all multi-indices $\alpha, \beta \in \N^n$ there is a constant $C_{\alpha\beta}$ with
$$\abs{\partial_\xi^\alpha \partial_x^\beta a(x, \xi)} \le C_{\alpha\beta} \, (1 + \abs{\xi})^{m - \rho \abs{\alpha} + \delta \abs{\beta}} \quad \text{for all } x, \xi \in \R^n.$$
Here $\partial_\xi^\alpha$ and $\partial_x^\beta$ are iterated [[partial derivatives|09EGC6F8PT]] and $\abs{\alpha} = \alpha_1 + \cdots + \alpha_n$.
The elements of $S^m_{\rho, \delta}$ are *symbols of order $m$* and *type $(\rho, \delta)$*.
One writes $S^m = S^m_{1, 0}$ and $S^{-\infty} = \bigcap_{m \in \R} S^m$.

## 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 $S^{m - \rho\abs{\alpha}}_{\rho, \delta}$.