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
$$ = \int_{-\infty}^{\infty} g(au) e^{-i 2 \pi
l u}du$$
$$ = \int_{-\infty}^{\infty} g(au) e^{-i 2 \pi [(l/a) (au)]}
\,{d(au) \over \vert a \vert}$$
$$ = {1 \over \vert a \vert}
\int_{-\infty}^{\infty} g(u) e^{- i 2 \pi (l/a) u} du$$
$$ = {1 \over
\vert a \vert} f\biggl({l \over a}\biggr) \qquad\qquad {\rm QED}$$