Commutativity of convolution

Remarks

The proof is the substitution $y = x - z$. Lebesgue measure is invariant under the reflection $z \mapsto -z$ and under translations, so $$\int_{\R^n} f(y) g(x - y) \, dy = \int_{\R^n} f(x - z) g(z) \, dz = (g * f)(x),$$ and the same computation with absolute values gives the equivalence of the two convergence conditions.

Example: on $\R$, with $f = \mathbf{1}_{[0, 1]}$ and $g = \mathbf{1}_{[0, 2]}$, both $f * g$ and $g * f$ give at $x$ the length of $[0, 1] \cap [x - 2, x]$ and of $[0, 2] \cap [x - 1, x]$, which are equal: both are $\min(x, 1)$ for $0 \le x \le 2$.

The statement uses that the group $\R^n$ is abelian. On a non-abelian locally compact group, convolution $f * g$ with $(f * g)(x) = \int f(y) g(y^{-1} x) \, dy$ is in general not commutative.

Going deeper

For a student

Commutativity, together with associativity and bilinearity, makes $L^1(\R^n)$ a commutative algebra under convolution. It is the reason why the convolution on the line can be written with either function shifted. Under the Fourier transform it becomes the commutativity of the pointwise product $\hat f \hat g = \hat g \hat f$.