Convolution: $ f (-)*g = g(-)* f$ does this mean both $f$ and $g$ have to be even functions?

Question

Assuming $f$ and $g$ are different functions, does $ f (-)*g = g(-)* f$ mean both $f$ and $g$ have to be even functions? In fact, this is equivalent to $f\star g = g \star f$ (i.e., cross-correlation of $f$ on $g$ equals to cross-correlation of $g$ on $f$). However, cross-correlation is in general non-commutative. So my guess is that for this equality to be hold, both $f$ and $g$ have to be even function. Is this right? What will be the formal proof for this? Any suggestions? Thanks for your time in advance.

Answer 1

For real-valued functions on $\mathbb{R}$ a simple inspection of the definition shows that $(f\star g)(x)=(g\star f)(-x)$, so if we impose commutativity we get that $f\star g$ must be an even function. This is the case for instance when one of the two functions is odd and the other is even, for if say $f$ is even and $g$ is odd then:

\begin{aligned} (f\star g)(x)\; & = \int f(u)g(u-x)\mathrm{d}u\\ & = \int f(-u) g(-u+x)\mathrm{d}(-u)\\ & = (f\star g)(-x)\;. \end{aligned}

Notice that if both $f$ and $g$ are even then $f\star g$ is even:

\begin{aligned} (f\star g)(x)\; & = \int_{-\infty}^\infty f(u)g(x-u)\mathrm{d}u\\ \text{Let }t=-u.\text{ Then }\mathrm{d}t=-\mathrm{d}u\\ & = -\int_\infty^{-\infty} f(-t) g(x+t)\mathrm{d}(t)\\ & = \int_{-\infty}^\infty f(t) g(-x-t)\mathrm{d}t\\ & = (f\star g)(-x)\;. \end{aligned}