Convolution product and zero-product property

Question

The convolution product for continuous integrable functions on $[0,+\infty)$ is defined as $$ (f * g)(t) = \int_0^t f(s) g(t-s) ds. $$ Does it has a zero-product property?
The paper Variational principles for linear elastodynamics by M. Gurtin in eq. $(2.7)$ says $$ \text{(b)} \quad \vartheta * \omega = 0 \quad \text{implies either } \vartheta = 0 \text{ or } \omega = 0; $$ but the book that he references in the footnote (J. Mikusinski - Operational Calculus - 1959) does not report such a property.

Somebody able to prove or provide a counterexample?

Answer 1

The simplest proof is probably to use the convolution theorem for the Laplace transform: For all $s>0$,

\begin{align} \int_0^{+\infty}e^{-st}(f*g)(t)\,dt=\Big(\int_0^{+\infty}e^{-st}f(t)\,dt\Big)\Big(\int_0^{+\infty}e^{-st}g(t)\,dt\Big)\,. \end{align} When $f,g\ge 0$ and $(f*g)\equiv 0$ then (since $f,g$ are both continuous) we must have either $f\equiv 0$ or $g\equiv 0$.

Answer 2

If $f,g$ are continuous functions defined on $[0,+\infty)$ and $\mathcal{L}$-transformable, then $$ f * g = 0 \qquad \implies \qquad f = 0 \lor g = 0 $$ Proof:
Set $F = \mathcal{L}[f], G = \mathcal{L}[g]$, then $$ \mathcal{L}[f * g] = F \cdot G = 0, $$ If exists $s_0$ such that $F(s_0) \neq 0$ then exists $\delta>0$ such that $F(s) \neq 0 \; \forall s:|s-s_0|<\delta$, of consequence $G(s)=0 \; \forall s:|s-s_0|<\delta$, but being a holomorphic function $G = 0$.