📂フーリエ解析

コンボリューションの性質

정리

畳み込みは次の性質を満たす。

(a) 交換法則

$$f \ast g = g \ast f$$

(b) 分配法則

$$f \ast (g+h) = f \ast g + f \ast h$$

(c) 結合法則

$$f \ast (g \ast h) = (f \ast g) \ast h$$

(d) スカラー乗の結合法則

$$a(f \ast g)=(af\ast g)=(f\ast ag)$$

(e) 微分

$$(f\ast g)^{\prime}=f^{\prime}\ast g=f\ast g^{\prime}$$

(f) 複素共役

$$\overline{f\ast g}=\overline{f} \ast \overline{g}$$

(g) ディラックのデルタ関数

$$f \ast \delta =f$$

証明

(a)

$$\begin{align*} f\ast g(x)&=\int _{-\infty} ^{\infty} f(y)g(x-y)dy \\ &=\int_{\infty} ^{-\infty} f(x-z)g(z)(-dz) \\ &=\int_{-\infty} ^{\infty} f(x-z)g(z)dz \\ &=\int_{-\infty} ^{\infty} g(z)f(x-z)dz \\ &=g\ast f(x) \end{align*}$$

2番目の等号で$x-y=z$に置き換えた。

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(b)

$$\begin{align*} f \ast (g+h)(x) &= \int f(y) (g+h)(x-y)dy \\ &= \int f(y)\left( g(x-y) + h(x-y) \right) dy \\ &= \int f(y)g(x-y)dy + \int f(y) h(x-y)dy \\ &= f\ast g(x)+f\ast h(x) \end{align*}$$

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(c)

$$\begin{align*} f \ast (g\ast h)(x) &= \int_{y} f(y) g\ast h(x-y)dy \\ &= \int_{y} f(y) h\ast g(x-y)dy \\ &= \int_{y} f(y) \int_{z} h(z)g(x-y-z)dzdy \\ &= \int_{z} h(z) \int_{y}f(y)g(x-z-y)dydz \\ &= \int_{z} h(z) f\ast g(x-z) dz \\ &=h \ast (f\ast g)(x) \\ &= (f\ast g)\ast h(x) \end{align*}$$

(d)

$$\begin{align*} a(f\ast g)(x) &= a\int f(y)g(x-y)dy \\ &= \int af(y) g(x-y)dy =af\ast g(x) \\ &=\int f(y)ag(x-y)dy=f\ast ag(x) \end{align*}$$ \ast \eta

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(e)

$$\begin{align*} (f\ast g)^{\prime}(x) &= \lim \limits_{h \to 0}\frac{f\ast g(x+h)-f\ast g(x) }{h} \\ &= \lim \limits_{h \to 0}\frac{\int f(y)g(x+h-y)dy-\int f(y)g(x-y)dy }{h} \\ &= \lim \limits_{h \to 0}\int f(y)\frac{ g(x+h-y)-g(x-y) }{h}dy \\ &=\int f(y)\lim \limits_{h \to 0}\frac{ g(x+h-y)-g(x-y) }{h}dy \\ &= \int f(y)g^{\prime}(x-y)dy \\ &= f\ast g^{\prime}(x) \end{align*}$$

この時、(a) によって

$$(f\ast g)^{\prime}(x)=(g\ast f)^{\prime}(x)=g\ast f^{\prime}(x)=f^{\prime}\ast g(x)$$

であるから、

$$(f\ast g)^{\prime}(x)=f^{\prime}\ast g(x)=f\ast g^{\prime}(x)$$

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(f)

$$\begin{align*} \overline{f\ast g(x)} &= \overline{\int f(y)g(x-y)dy} \\ &= \int \overline{f(y)g(x-y)}dy \\ &= \int \overline{f(y)}\ \overline{g(x-y)}dy \\ &= \overline{f} \ast \overline{g}(x) \end{align*}$$

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(g)

$$\begin{align*} f \ast \delta (x) &= \int f(y)\delta (x-y)dy \\ &= f(x)\int \delta (x-y)dy \\ &= f(x) \end{align*}$$

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