Convolution ring has no zero divisors

Remarks

Commutativity and associativity follow from the corresponding facts on the line, and distributivity is linearity of the integral. For the absence of zero divisors, suppose $f * g = 0$ and $f(t_0) \ne 0$. By continuity $f$ is nonzero near $t_0$, so $f$ does not vanish almost everywhere on $(0, \lambda)$ for any $\lambda > t_0$. For each $T > t_0$, the Titchmarsh convolution theorem then gives $g = 0$ almost everywhere on $(0, T - t_0)$; letting $T \to \infty$ and using continuity gives $g = 0$.

$\mathcal{C}$ has no unit element. If $e * f = f$ for all $f$, then $e * 1 = 1$, but $(e * 1)(0) = 0 \ne 1$.

Example: $1 * 1 = t$ and $t * t = t^3/6$, and in general no product of nonzero polynomials in $\mathcal{C}$ is $0$, in line with the formula $t^m * t^n = \frac{m!\,n!}{(m+n+1)!} t^{m+n+1}$.

Going deeper

For a student

A commutative rng without zero divisors has a field of fractions, built from pairs $f/g$ as for the rational numbers. For $\mathcal{C}$ this is the Mikusiński field of operators, the setting of the Mikusiński operational calculus. The same conclusion holds for locally integrable functions, with equality almost everywhere.