Duality

Exploration

By exploring the similarity in the formulas for the Fourier transform and its inverse:

\[ \mathcal{F} f(s) = \int_{-\infty}^{+\infty} e^{-2 \pi i s t} f(t) dt \]
\[ \mathcal{F}^{-1} f(t) = \int_{-\infty}^{+\infty} e^{2 \pi i s t} f(s) ds \]
\[ \text{Let } f^- (x) = f(-x) \]
\[\begin{split} \therefore (\mathcal{F} f)^- (x) & = \int_{-\infty}^{+\infty} e^{-2 \pi i \cdot (-x) \cdot t} f(t) dt \\ & = \int_{-\infty}^{+\infty} e^{2 \pi i x t} f(t) dt \\ & = \mathcal{F}^{-1} f(x) \end{split}\]
\[ \mathcal{F} f^- (x) = \int_{-\infty}^{+\infty} e^{-2 \pi i s t} f(-t) dt \]
\[ \text{Let } u = -t \implies t = -u \]
\[\begin{split} \therefore \mathcal{F} f^- (x) & = \int_{+\infty}^{-\infty} e^{-2 \pi i s \cdot (-u)} f(u) d(-u) \\ & = \int_{-\infty}^{+\infty} e^{2 \pi i s u} f(u) du \\ & = \mathcal{F}^{-1} f (x) \end{split}\]

Formulas

\[ (\mathcal{F} f)^- = \mathcal{F}^{-1} f \]
\[ \mathcal{F} f^- = \mathcal{F}^{-1} f \]
\[ \therefore (\mathcal{F} f)^- = \mathcal{F} f^- \]
\[ \text{Let } g^- = f \iff g = f^- \]
\[\begin{split} \therefore \mathcal{F} \mathcal{F} f & = \mathcal{F} \mathcal{F} g^- \\ & = \mathcal{F} \mathcal{F}^{-1} g \\ & = g \\ & = f^- \end{split}\]

Application

\[ \DeclareMathOperator{\sinc}{sinc} \mathcal{F} \sinc = \pi \]

Proof

\[ \because \mathcal{F} \pi (x) = \sinc (x) \]
\[\begin{split} \therefore \mathcal{F} \sinc & = \mathcal{F} \mathcal{F} \pi \\ & = \pi^- \\ & = \pi \end{split}\]
\[ \tag*{$\blacksquare$} \]

Back to Fourier Transform.