# Convolution Is Commutative

22 Oct 2013The convolution of two functions f and g is written as \(f\ast g\). \[\begin{align*} \left(f\ast g\right)(t)&:=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)\, d\tau\\&=\int_{-\infty}^{\infty}f(t-\tau)g(\tau)\, d\tau\\&=\left(g\ast f\right)(t) \end{align*}\]

Convolution appears quite often when one deals with differential equations and Fourier transforms. Interestingly, the convolution is commutative. That is, \(f\ast g=g\ast f \). This fact bothers me quite a bit as, at the first glance, the definition does not suggest so. Here is my attempt to show that it is indeed commutative to convince myself.

Let \(u:=t-\tau\). So, \(du=-d\tau\). \[\begin{align*} \left(f\ast g\right)(t)&:=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)\, d\tau\\&=\int_{u(\tau=-\infty)}^{u(\tau=\infty)}f(t-u)g(u)(-du)\\&=-\int_{\infty}^{-\infty}g(u)f(t-u)\, du\\&=\int_{-\infty}^{\infty}g(u)f(t-u)\, du\\&=\left(g\ast f\right)(t) \end{align*}\]