$\exp(-x^2)$ 是一个Schwartz函数

设

$$\frac{\ud^{n}}{\ud x^{n}}\exp\left(-x^{2}\right)=p_{n}(x)\exp\left(-x^{2}\right),$$

则有$p_{0}(x)=1$, $p_{1}(x)=-2x$. 计算

$$\begin{align*} \frac{\ud^{n+1}}{\ud x^{n+1}}\exp\left(-x^{2}\right) & =\frac{\ud}{\ud x}\frac{\ud^{n}}{\ud x^{n}}\exp\left(-x^{2}\right)\\ & =\frac{\ud}{\ud x}\left(p_{n}(x)\exp\left(-x^{2}\right)\right)\\ & =p'_{n}(x)\exp\left(-x^{2}\right)-2xp_{n}(x)\exp(-x^{2})\\ & =\left(p_{n}'(x)-2xp_{n}(x)\right)\exp(-x^{2}). \end{align*}$$

所以我们期望有关系

$$p_{n+1}(x):= p'_{n}(x)-2xp_{n}(x)$$

成立. 容易证明对于任何$n$, $p_{n}(x)$是$x$的多项式.

考虑$\exp(x^{2})$的Maclaurin级数有

$$\exp(x^{2})\ge\frac{\left|x\right|^{k}}{k!}\left|x\right|^{k}\Longrightarrow\left|x\right|^{-k}k!\ge\left|x\right|^{k}\exp(-x^{2}).$$

所以

$$\left|x^{k}\exp(-x^{2})\right|\le\frac{k!}{\left|x\right|^{k}}\le k!,\qquad\left|x\right|\ge1.$$ $$\left|x^{k}\exp(-x^{2})\right|\le1,\qquad\left|x\right|\le1.$$

即对于任何$k\in\NN$, 有

$$\sup_{x}\left|x^{k}\exp(-x^{2})\right|\le k!<\infty,$$

即得$\exp(-x^{2})\in\mathcal{S}(\RR)$.