Why does convolution become multiplication when taking the Fourier transform? Why is it, that if I have a convolution and I take the Fourier Transform, it becomes a multiplication?
 A: Basically, this is because Fourier modes are eigenfunctions of translation operators: 
$$\underbrace{e^{ik(x-p)}}_{\text{translated function}} = \underbrace{e^{-ipk}}_{\text{scalar}} ~ \underbrace{e^{ikx}}_{\text{original function}}.$$A convolution operator is build up from sums or integrals of translation operators, and hence Fourier modes also diagonalize convolution operators. The function-space analog of a diagonal matrix is a multiplication operator.
Making this precise requires some functional analysis, but this is the basic idea and the correct intuition.
A: Assuming everything converges absolutely, with the change of variable $y = t-x$ :
$$\mathcal{F}[g(x)](\xi)\ \mathcal{F}[h(x)](\xi) = \left(\int_{-\infty}^\infty g(x)  e^{-2i \pi\xi x}dx\right)\left( \int_{-\infty}^\infty h(x)  e^{-2i \pi\xi x}dx\right)$$ $$= \int_{-\infty}^\infty \int_{-\infty}^\infty g(x) h(y) e^{-2i \pi\xi x}e^{-2i \pi\xi y}dxdy= \int_{-\infty}^\infty \int_{-\infty}^\infty g(x) h(t-x) e^{-2i \pi\xi x}e^{-2i \pi\xi (t-x)}dxdt$$ $$ = \int_{-\infty}^\infty \left(\int_{-\infty}^\infty g(x) h(t-x) dx\right)e^{-2i \pi\xi t} dt=\mathcal{F}[g\ast h(x)](\xi)$$
The opposite direction needs the Fourier inversion theorem, with $G(\xi) = \mathcal{F}[g(x)](\xi), H(\xi ) =\mathcal{F}[h(x)](\xi)$, and since the inverse Fourier transform is essentially the same operator as the Fourier transform, it means that
$$g \ast h(x) =  \mathcal{F}^{-1}[G(\xi)H(\xi)](x) \quad \implies \quad G \ast H(\xi) =  \mathcal{F}[g(x) h(x)](\xi)$$
A: The Fourier transform is the decompostion of a signal as a sum of sinuoids.
When you convolve two such sums, by the superposition principle you get a double sum of convolved sinusoids. By orthogonality, the individual convolutions amount to Dirac deltas (modulated by the phase shift). Hence, the Fourier decomposition of a convolution is given by the product of the coefficients where the frequencies match.
Informally, for every component of the first signal
$$a_ie^{i\omega t}*\sum_j b_je^{j\omega t}=\sum_j a_i b_j(e^{i\omega t}*e^{j\omega t})=\sum_j a_ib_j\delta_{ij}=a_ib_i.$$
(Summations are understood in a broad sense.)
