$f(x) \ge 0$, $g(x) \ge 0$ are defined on $[0,\infty)$ and $f(x) \sim x^{-a}, \ x \to \infty$, where $a>1$. The integrals $\int_0^\infty f(x)dx<\infty$ and $\int_0^\infty g(x) dx<\infty$.
Is it true that $\int_0^x f(x-u)g(u)du \sim x^{-a}\int_0^\infty g(x) dx, x \to \infty?$
Not without extra hypotheses on $g$, or maybe on $f$. The convolution is larger than $$\int_{x-1}^x f(x-t)g(t)\,dt=\int_0^1f(t)g(x-t)\,dt.$$ If, say, $f$ is bounded then that tends to $0$ as $x\to\infty$, but it can tend to $0$ slower than $x^{-a}$.