Should $f(x) \equiv 0$ if $0\le f'(x)\le f(x)$ and $f(0)=0$? Assume $f(x)$ is a real-function defined on $[0,+\infty)$ and satisfies the followings:


*

*$f'(x) \geq 0$

*$f(0)=0$

*$f'(x) \leq f(x)$
Should we always have $f(x) \equiv 0$ ? Thanks for any solution.

As $f(0)=0$ and for all $x\geq0$, $f'(x)\geq0$, we have $f(x)\geq0, \forall x\geq0$.
Define
$$g(x)=\mathrm e^{-x}f(x)$$
and compute 
$$g'(x)=\mathrm e^{-x}\left(f'(x)-f(x)\right)\leq 0$$
as $g(0)=0$ we have $g(x)\leq 0$ for all $x\geq0$.
Therefore we have
$$f(x)=\mathrm e^xg(x)\leq 0,\quad \forall x\geq0,$$
which proves $f\equiv0$.
