用 Fourier 变换证明 Leibniz 法则

证明. 利用傅里叶变换的性质,记傅里叶变换算子为 $\mathcal{F}$ ,且注意到:
$$
\mathcal{F}\{f \cdot g\} = \frac{1}{2\pi} \mathcal{F}\{f\} * \mathcal{F}\{g\}, \quad (j\omega)^{n} =\sum_{k=0}^{n} \binom{n}{k} (j\eta)^{n-k} (j(\omega - \eta))^{k}.
$$

首先,对 $\mathcal{D}^{n}(fg)$ 进行傅里叶变换:
$$
\mathcal{F}\{\mathcal{D}^{n}(fg)\} = (j\omega)^{n} \mathcal{F}\{fg\}.
$$

代入卷积公式:
$$
= \frac{1}{2\pi} (j\omega)^{n} \left( \mathcal{F}\{f\} * \mathcal{F}\{g\} \right)
= \frac{1}{2\pi} (j\omega)^{n} \int_{\mathbb{R}} \mathcal{F}\{f\}(\eta) \mathcal{F}\{g\}(\omega - \eta) \, \mathrm{d}\eta.
$$

将 $(j\omega)^{n}$ 展开:
$$
= \frac{1}{2\pi} \sum_{k=0}^{n} \binom{n}{k} \int_{\mathbb{R}} (j\eta)^{n-k} \mathcal{F}\{f\}(\eta) \cdot [j(\omega - \eta)]^{k} \mathcal{F}\{g\}(\omega - \eta) \, \mathrm{d}\eta.
$$

注意到:
$$
(j\eta)^{n-k} \mathcal{F}\{f\}(\eta) = \mathcal{F}\{\mathcal{D}^{n-k}f\}(\eta), \quad [j(\omega - \eta)]^{k} \mathcal{F}\{g\}(\omega - \eta) = \mathcal{F}\{\mathcal{D}^{k}g\}(\omega - \eta),
$$
因此:
$$
= \frac{1}{2\pi} \sum_{k=0}^{n} \binom{n}{k} \int_{\mathbb{R}} \mathcal{F}\{\mathcal{D}^{n-k}f\}(\eta) \cdot \mathcal{F}\{\mathcal{D}^{k}g\}(\omega - \eta) \, \mathrm{d}\eta
= \sum_{k=0}^{n} \binom{n}{k} \mathcal{F}\left\{ \mathcal{D}^{n-k}f \cdot \mathcal{D}^{k}g \right\}.
$$

两边同时进行傅里叶逆变换,即得:
$$
\mathcal{D}^{n}(fg) = \sum_{k=0}^{n} \binom{n}{k} \mathcal{D}^{n-k}f \cdot \mathcal{D}^{k}g.
$$