Convolution theorem

Remarks

The proof is one application of Fubini's theorem. Write $e^{-2\pi i x \cdot \xi} = e^{-2\pi i y \cdot \xi} e^{-2\pi i (x - y) \cdot \xi}$ inside the double integral of $f(y) g(x - y) e^{-2\pi i x \cdot \xi}$. The double integral converges absolutely because $\int\int \abs{f(y)} \abs{g(x - y)}\, dy\, dx = \norm{f}_1 \norm{g}_1$. Integrating first in $x$ and then in $y$ gives $\hat f(\xi) \hat g(\xi)$.

Example: on $\R$, let $f = g = \mathbf{1}_{[-1/2, 1/2]}$. Then $\hat f(\xi) = \sin(\pi \xi)/(\pi \xi)$, and $f * g$ is the triangle function $\max(1 - \abs{x}, 0)$. So the Fourier transform of the triangle function is $\sin^2(\pi \xi)/(\pi \xi)^2$.

There is a dual form: for $f, g$ in the Schwartz space, $\widehat{f g} = \hat f * \hat g$. The same identity holds for Fourier series, where the convolution on the circle becomes the product of Fourier coefficients.

Going deeper

For a student

The theorem is why the Fourier transform solves linear equations with constant coefficients: such equations are convolution equations, and the transform makes them algebraic. It also shows that $L^1(\R^n)$ with convolution is a commutative algebra, and that $f \mapsto \hat f$ is an algebra homomorphism into bounded continuous functions with pointwise product. Fast convolution of long sequences uses the discrete version of the theorem together with the fast Fourier transform.