Duality¶
Exploration¶
By exploring the similarity in the formulas for the Fourier transform and its inverse:
\[ \begin{align}\begin{aligned}\mathcal{F} f(s) =
\int_{-\infty}^{+\infty} e^{-2 \pi i s t} f(t) dt\\\mathcal{F}^{-1} f(t) =
\int_{-\infty}^{+\infty} e^{2 \pi i s t} f(s) ds\\\begin{split}\\\end{split}\\\text{Let } f^- (x) = f(-x)\\\begin{split}\\\end{split}\\\begin{split}\therefore
(\mathcal{F} f)^- (x) & =
\int_{-\infty}^{+\infty} e^{-2 \pi i \cdot (-x) \cdot t} f(t) dt
\\ & =
\int_{-\infty}^{+\infty} e^{2 \pi i x t} f(t) dt
\\ & =
\mathcal{F}^{-1} f(x)\end{split}\\\begin{split}\\\end{split}\\\mathcal{F} f^- (x) =
\int_{-\infty}^{+\infty} e^{-2 \pi i s t} f(-t) dt\\\begin{split}\\\end{split}\\\text{Let } u = -t \implies t = -u\\\begin{split}\\\end{split}\\\begin{split}\therefore
\mathcal{F} f^- (x) & =
\int_{+\infty}^{-\infty} e^{-2 \pi i s \cdot (-u)} f(u) d(-u)
\\ & =
\int_{-\infty}^{+\infty} e^{2 \pi i s u} f(u) du
\\ & =
\mathcal{F}^{-1} f (x)\end{split}\end{aligned}\end{align} \]
Formulas¶
\[ \begin{align}\begin{aligned}(\mathcal{F} f)^- = \mathcal{F}^{-1} f\\\mathcal{F} f^- = \mathcal{F}^{-1} f\\\therefore
(\mathcal{F} f)^- = \mathcal{F} f^-\\\begin{split}\\\end{split}\\\text{Let } g^- = f \iff g = f^-\\\begin{split}\\\end{split}\\\begin{split}\therefore
\mathcal{F} \mathcal{F} f & =
\mathcal{F} \mathcal{F} g^-
\\ & =
\mathcal{F} \mathcal{F}^{-1} g
\\ & =
g
\\ & =
f^-\end{split}\end{aligned}\end{align} \]
Application¶
\[ \begin{align}\begin{aligned}\DeclareMathOperator{\sinc}{sinc}\\\mathcal{F} \sinc = \pi\end{aligned}\end{align} \]
Proof
\[ \begin{align}\begin{aligned}\because
\mathcal{F} \pi (x) = \sinc (x)\\\begin{split}\\\end{split}\\\begin{split}\therefore
\mathcal{F} \sinc & =
\mathcal{F} \mathcal{F} \pi
\\ & =
\pi^-
\\ & =
\pi\end{split}\end{aligned}\end{align} \]
\[\tag*{$\blacksquare$}\]
Back to Fourier Transform.