合成積のFourier変換

$ {\cal F}(x*y)(\omega)=\sqrt{2\pi}{\cal F}(x)(\omega){\cal F}(y)(\omega)

証明

$ {\cal F}(x*y)(\omega)=\frac1{\sqrt{2\pi}}\int_{(t,\tau)\in\R^2}x(\tau)y(t-\tau)e^{-i\omega t}\mathrm d\tau\mathrm dt

$ =\frac1{\sqrt{2\pi}}\int_{(\tau,t-\tau)\in\R^2}x(\tau)y(t-\tau)e^{-i\omega((t-\tau)+\tau)}\mathrm d\tau\mathrm d(t-\tau)

$ =\frac1{\sqrt{2\pi}}\int_{(\tau,t)\in\R^2}x(\tau)y(t)e^{-i\omega(t+\tau)}\mathrm d\tau\mathrm d t

$ =\frac1{\sqrt{2\pi}}\int_{(\tau,t)\in\R^2}x(\tau)e^{-i\omega\tau}y(t)e^{-i\omega t}\mathrm d\tau\mathrm d t

$ = \sqrt{2\pi}\frac1{\sqrt{2\pi}}\int_\R x(\tau)e^{-i\omega\tau}\mathrm d\tau\frac1{\sqrt{2\pi}}\int_\R y(t)e^{-i\omega t}\mathrm dt

$ =\sqrt{2\pi}{\cal F}(x)(\omega){\cal F}(y)(\omega)