Let $a>0$ and $g$ be a non-negative, smooth function on $[0,+\infty[$ such that $g(0)=0$ and $\lim_{t\to\infty} g(t)=0$. Consider the integral, $$ I(t) = \int_{0}^t e^{a(s-t)}g(s)\mathrm{d}s.$$ I would like to obtain an upper-bound on $I(t)$, holding for all $t\geq 0$ and that vanishes as $t\to\infty$. I think a bound of the form $I(t)\leq \max(C e^{-at},g(t))$ can be obtained, for some $C>0$, but so far I did not succeed.
What I tried: Defining $h(t)=e^{-at}$ and extending $g$ for all $t<0$ by $g(t)=0$, we can express $I$ as a convolution: $$ I(t) = \int_{0}^t h(t-s)g(s)\mathrm{d}s = (h\ast g)(t) - \int_{t}^{+\infty} h(t-s) g(s)\mathrm{d}s.$$ So it could be sufficient to bound $(h\ast g)(t)$. I checked this related post but the result is only asymptotic (for large $t$).
Let $$M=\max_{t\ge 0}g(t),\quad G(t)=\max_{ s\ge t}g(s)$$ Then $$I(t)=e^{-at}\int\limits_0^{t/2}e^{as}g(s)\,ds +e^{-at}\int\limits_{t/2}^te^{as}g(s)\,ds \\ \le {M\over a}e^{-at/2}+ {1\over a}G(t/2) $$ The last expression tends to $0$ at $\infty.$