合成積のFourier変換

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

証明

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

$ =\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)

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

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

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

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