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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?

Vincenzo Tibullo
  • 10,955

2 Answers2

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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 $G = 0$.

Vincenzo Tibullo
  • 10,955
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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 $g\equiv 0$.

Kurt G.
  • 14,198
  • If $F,G$ are the two Laplace transform, $F(s)G(s)=0$ is valid pointwise, but does not mean that one of the two vanish identically, I think – Vincenzo Tibullo May 04 '22 at 11:46
  • $F$ is an integral over a nonnegative function $f$. When $F(s)$ is zero for a single $s$ that means the continuous function $f$ has to vanish identically. Likewise this holds for $G$. – Kurt G. May 04 '22 at 12:46
  • OK, true, but the hypothesis $f,g>0$ is too strong, and not in my initial question. – Vincenzo Tibullo May 04 '22 at 12:54
  • Correct. This indicates what counter examples one should look for. – Kurt G. May 04 '22 at 13:05
  • @enzotib . Here is a proof that shows the result without the assumption $f,g\ge 0$ as long as $f,g$ are piecewise continuous. That's the case in your question. Please dive into this. I don't have time for everything. – Kurt G. May 04 '22 at 18:59