$f$ is differentiable on $[a,b]$, $f'(x) \leq A|f(x)|$ where $A$ is a non-negative constant.
If $f(a)=0$ show $f(x)=0, \forall x\in [a,b]$
I imagine the proof uses the Mean Value Theorem but I have not been able to get it to work.
I know $|f(x)|=|f'(c)|(x-a)$ where $c \in [a,x]$, so $|f(x)| \leq A\ |f(c)|(x-a)$ where $c\leq x$ And I guess I could sort of iterate this to keep getting a smaller $c$ but I don't see why it must go all the way to zero.