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.