# 卷积定理

$\mathcal{F}\{f*g\} = \mathcal{F}\{f\} \cdot \mathcal{F}\{g\}$

$\mathcal{F}\{f \cdot g\}= \mathcal{F}\{f\}*\mathcal{F}\{g\}$

$f*g= \mathcal{F}^{-1}\big\{\mathcal{F}\{f\}\cdot\mathcal{F}\{g\}\big\}$

## 证明

fg属于L1(Rn)。$F$$f$的傅里叶变换，$G$$g$的傅里叶变换：

$F(\nu) = \mathcal{F}\{f\} = \int_{\mathbb{R}^n} f(x) e^{-2 \pi i x\cdot\nu} \, \mathrm{d}x$
$G(\nu) = \mathcal{F}\{g\} = \int_{\mathbb{R}^n}g(x) e^{-2 \pi i x\cdot\nu} \, \mathrm{d}x,$

$h(z) = \int\limits_{\mathbb{R}} f(x) g(z-x)\, \mathrm{d} x.$

$\int\!\!\int |f(z)g(x-z)|\,dx\,dz=\int |f(z)| \int |g(z-x)|\,dx\,dz = \int |f(z)|\,\|g\|_1\,dz=\|f\|_1 \|g\|_1.$

\begin{align} H(\nu) = \mathcal{F}\{h\} &= \int_{\mathbb{R}} h(z) e^{-2 \pi i z\cdot\nu}\, dz \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}^n} f(x) g(z-x)\, dx\, e^{-2 \pi i z\cdot \nu}\, dz. \end{align}

$H(\nu) = \int_{\mathbb{R}} f(x)\left(\int_{\mathbb{R}^n} g(z-x)e^{-2 \pi i z\cdot \nu}\,dz\right)\,dx.$

$H(\nu) = \int_{\mathbb{R}} f(x) \left( \int_{\mathbb{R}} g(y) e^{-2 \pi i (y+x)\cdot\nu}\,dy \right) \,dx$
$=\int_{\mathbb{R}} f(x)e^{-2\pi i x\cdot \nu} \left( \int_{\mathbb{R}} g(y) e^{-2 \pi i y\cdot\nu}\,dy \right) \,dx$
$=\int_{\mathbb{R}} f(x)e^{-2\pi i x\cdot \nu}\,dx \int_{\mathbb{R}} g(y) e^{-2 \pi i y\cdot\nu}\,dy.$

$H(\nu) = F(\nu) \cdot G(\nu),$

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