The convolution $h$ of two functions $f$ and $g$ is defined by
$$h(x) \equiv \int_{-\infty}^\infty f(u) g(x - u) du$$
and the convolution operator is often indicated by the symbol $\otimes$, so the above equation can be written
$$h = f \otimes g$$

Convolution Theorem:

If the Fourier transform of $f(x)$ is $F(s)$ and the Fourier transform of $g(x)$ is $G(s)$, then the Fourier transform $H(s)$ of their convolution $h(x) = f \otimes g$ is simply the product of the Fourier transforms of $f$ and $g$:
$$H(s) = F(s) \cdot G(s )$$

Proof:

By definition, the Fourier transform of $h(x)$ is
$$H(s) \equiv \int_{-\infty}^\infty h(x) e^{- i 2 \pi s x} d x$$
By the definition of convolution, the convolution $h = f \otimes g$ is
$$H(s) = \int_{- \infty}^\infty \biggl[ \int_{-\infty}^\infty f(u) g(x-u) du \biggr] e^{ - i 2 \pi s x} d x$$
$$H(s) = \int_{-\infty}^\infty \int_{-\infty}^\infty f(u) g(x - u) e^{- i 2 \pi s x } d x d u$$
$$H(s) = \int_{-\infty}^\infty f(u) \biggl[ \int_{ -\infty}^\infty g(x - u) e^{- i 2 \pi s x} d x \biggr] du$$
Replace $x$ by $x' \equiv (x - u)$ within the integration over $dx$:
$$H(s) = \int_{-\infty}^\infty f(u) \biggl[ \int_{-\infty}^\infty g(x - u) e^{- i 2 \pi s (x - u)} e^{ - i 2 \pi s u} d (x - u ) \biggr] d u$$
$$H(s) = \int_{-\infty}^\infty f (u) e^{ - i 2 \pi s u} \biggl[ \int_{-\infty}^\infty g(x') e^{- i 2 \pi s x'} d x' \biggr] d u$$
The quantity in square brackets above is just $G(s)$, the Fourier transform of $g(x)$, so
$$H(s) = \biggl[ \int_{-\infty}^\infty f(u) e^{- i 2 \pi s u} du \biggr] G(s)$$ Likewise, the quantity in square brackets above is $F(s)$, so
$$H(s) = F(s) \cdot G(s) \qquad \qquad {\rm QED}$$