prove that the following function is: $f(x) = 0$ 
let $f: [0,1] \to \mathbb R$ , $f$ is differentiable
$f(0) = 0$
$|f'(x)|\le|f(x)|$ for $x\in [0,1]$
prove that $f(x) =0$ for $x\in [0,1]$

i believe that i need to somehow use the mean value theorem iteratively
any hints?

Choose $x \in [0,1]$. By the mean-value theorem,
$$f(x) = f(x) - f(0) = f'(c)(x-0) = f'(c)x$$
for some $c \in (0,x)$. But
$$|f'(c)x| \leq |f(c)||x| = |f(c) - f(0)||x| \leq |f'(d)||x||c| \leq |f'(d)||x|^2$$
for some $d \in (0,c)$. Do you see how to iterate this argument?

More hints:

 Note that $f$ is continuous and hence bounded on the compact set $[0,1]$, and by the hypothesis that $|f'| \leq |f|$, $f'$ is also bounded.



 Iterate the above to show $|f(x)| \leq M |x|^n$ for some absolute constant $M$ and all $n$.



 Conclude that $f(x) = 0$ for all $x \in [0,1)$. To obtain the result at $1$, either use the continuity of $f(x)$, or the following argument: you now know that $f(0.5) = 0$, so repeat the above argument, but center it at $0.5$ instead of $0$. You should get some bound $f(1) \leq M |1 - 0.5|^n$. 

