Fourier transform property: Duality

Statement: if $g(t)\leftrightarrow G(f)$, then the function $G$ used as a time signal has the transform

$$G(t)\leftrightarrow g(-f)$$

Explanation: the forward and inverse Fourier integrals differ only in the sign of the exponent. Swapping the roles of $t$ and $f$ in the inverse transform $g(t)=\int G(f)e^{j2\pi ft}df$ and replacing $t$ by $-f$ gives $g(-f)=\int G(t)e^{-j2\pi ft}dt$. For even $g(t)$, $G(t)\leftrightarrow g(f)$.

Example: since $\text{rect}(t/T)\leftrightarrow T\,\text{sinc}(fT)$, duality gives $T\,\text{sinc}(tT)\leftrightarrow\text{rect}(f/T)$; a sinc pulse in time has a rectangular (ideally band-limited) spectrum. Likewise $\delta(t)\leftrightarrow1$ gives $1\leftrightarrow\delta(f)$. The property halves the number of transform pairs one must remember.