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.
 A: To add to the very nice techniques already used, here's another way of seeing that $f = 0$. Put $M_x = \sup_{t\in[0,x]}{|f(t)|}$. (Actually, as @V.Rosetto notes, $f$ is increasing and so $M_x = f(x)$, which simplifies things!) For each $x>0$, the mean value theorem furnishes a $t\in(0,x)$ such that $f(x) = xf'(t)$. But then
\begin{align*}
M_x = \sup_{t\in[0,x]}|f(t)| \leq \sup_{t\in[0,x]} t|f'(t)| \leq \sup_{t\in[0,x]} t|f(t)|\leq xM_x \tag{1}
\end{align*}
for $x>0$. Then for $x\in(0,1)$, we must have $M_x = 0$, lest $(1)$ lead to the contradictory inequality $M_x<M_x$. Thus $f = 0$ on $[0,1)$. 
But now $x\mapsto f(x-1)$ satisfies the same conditions as $f$ on $[0,\infty)$, and so it vanishes on $[0,1)$ as well. This means that $f$ vanishes on $[0,2)$. Now consider $x\mapsto f(x-2)$, and so on.
A: 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 $f\equiv0$.
A: Let $x_0\in[0,x]$ be a point in which $f'$ assumes a maximum. Then 
$$f(x) = f(x) - f(0) = \int_0^xf'(t)dt \le xf'(x_0) \le xf(x_0) \le xf(x)$$
so $f$ must be $0$ on $[0,1)$ and by continuity $f(1) = 0$. Since $f(x - 1)$ satisfies properties 1), 2), 3), the result follows by induction.
