F(F(f))(ω)=f(-ω)

証明

$ \mathcal F(f)(\omega)=\frac1{\sqrt{2\pi}}\int_\R f(t)e^{-i\omega t}\mathrm dt

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

$ =\mathcal F^{-1}(f)(-\omega)

$ \underline{\therefore\mathcal F^2(f)(\omega)=f(-\omega)\quad}_\blacksquare