Given a continuous function $g$, if the convolution $g*g(t)$ (defined as: $\int^t_0(g(r)g(t−r))dr$ equals $0$, $\forall t\geq 0$, then $g=0$.
My attempt was to use Laplace transformation, but such transformation for the function could be non-existent. I'm also aware of the Titchmarsh theorem, but all the proofs I've found involve material that I hadn't studied yet. The proof for the claim above should neither be very lengthy, nor include material that is studied in Harmonic Analysis.
How can one prove this claim using relatively simple means?
To elaborate on my comment, here's my thought process: Suppose $g\neq 0$ and let $X = \{ t \geqslant 0 \, \vert g(t) \neq 0\}$ which, by assumption, is nonempty. Denote by $t_\ast = \inf X$. Since $g$ is continuous there exists some $\varepsilon>0$ such that $g(\tau) \neq 0$ for all $\tau \in (t_\ast ,t_\ast +\varepsilon)$. Without loss of generality, suppose $g(\tau) > 0$ for all $\tau \in (t_\ast ,t_\ast +\varepsilon)$.
Now consider $(g\ast g)(2t_\ast +\varepsilon)$. Since $g(\tau)=0$ when $\tau \in (0,t_\ast)$, and $g(2t_\ast + \varepsilon-\tau) = 0$ when $\tau \in (\tau_\ast +\varepsilon , 2t_\ast +\varepsilon)$, it follows \begin{align*} (g\ast g)(2t_\ast +\varepsilon) &= \int_{t_\ast}^{t_\ast+\varepsilon} g(\tau) g(2t_\ast +\varepsilon -\tau) d \tau. \end{align*} But for $\tau \in (t_\ast,t_\ast+\varepsilon)$, $2t_\ast+\varepsilon-\tau \in (t_\ast , t_\ast + \varepsilon)$, so $g(2t_\ast +\varepsilon -\tau)>0$, so $$(g\ast g)(2t_\ast +\varepsilon)>0.$$