Fourier-Transformation/R^n/Funktion/Reziprozität/Fakt/Beweis

Beweis

Wir lassen den Vorfaktor in der Fourier-Transformierten weg. Es ist nach dem Satz von Fubini

{}{\begin{aligned}\int _{\mathbb {R} ^{n}}f{\hat {g}}d\lambda ^{n}&=\int _{\mathbb {R} ^{n}}f({\mathfrak {u}}){\hat {g}}({\mathfrak {u}})d{\mathfrak {u}}\\&={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}f({\mathfrak {u}})\cdot {\left(\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }g({\mathfrak {t}})d{\mathfrak {t}}\right)}d{\mathfrak {u}}\\&={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}\times \mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }\cdot f({\mathfrak {u}})\cdot g({\mathfrak {t}})d{\mathfrak {t}}d{\mathfrak {u}}\\&={\frac {1}{(2\pi )^{n/2}}}\int _{\mathbb {R} ^{n}}g({\mathfrak {t}})\cdot {\left(\int _{\mathbb {R} ^{n}}e^{-{\mathrm {i} }\left\langle {\mathfrak {u}},{\mathfrak {t}}\right\rangle }f({\mathfrak {u}})d{\mathfrak {u}}\right)}d{\mathfrak {t}}\\&=\int _{\mathbb {R} ^{n}}g({\mathfrak {t}})\cdot {\hat {f}}({\mathfrak {t}})d{\mathfrak {t}}\\&=\int _{\mathbb {R} ^{n}}{\hat {f}}gd\lambda ^{n}.\end{aligned}}