# The Similarity Theorem for Fourier Transforms:

If $f(l)$ is the Fourier transform of $g(u)$, then
$${ 1 \over \vert a \vert} f\biggl({l \over a}\biggr)$$
is the Fourier transform of $g(au)$, where $a \neq 0$ is a constant.

In words, the similarity theorem states that making a function $g$ wider or narrower makes its Fourier transform $f$ narrower and taller or wider and shorter, respectively, always conserving the area under the transform.

Proof: From the definition of a Fourier transform,
$$f(l) = \int_{-\infty}^{\infty} g(u) e^{-i 2 \pi l u} du ,$$
the Fourier transform of $g(au)$ is
$$= \int_{-\infty}^{\infty} g(au) e^{-i 2 \pi l u}du$$
$$= \int_{-\infty}^{\infty} g(au) e^{-i 2 \pi [(l/a) (au)]} \,{d(au) \over \vert a \vert}$$
$$= {1 \over \vert a \vert} \int_{-\infty}^{\infty} g(u) e^{- i 2 \pi (l/a) u} du$$
$$= {1 \over \vert a \vert} f\biggl({l \over a}\biggr) \qquad\qquad {\rm QED}$$