Solution
= Solution
Put $h(x)=|f(0)|+\int_0^xg(t)|f(t)|dt$. Then $|f|\le h$ and $h\prime\le gh$. If $h\gt 0$, $(\log h)\prime\le g$; integration gives $h(x)\le h(0)e^{\int_0^xg}$. For $f(0)=0$, apply the same argument after adding $\varepsilon$ and let $\varepsilon\downarrow0$.