TITLE: consequence of mean value theorem?
QUESTION [5 upvotes]: Suppose that $f:[0,1]\to\mathbb{R}$ is differentiable on (0,1) and continuous on [0,1]. Would like to assert that if f(0)=0, and $|f'(x)|\leq |f(x)|$ for each $x\in (0,1)$, then f is the zero function.  
I have tried applying several different variations of the mean value theorem, but nothing useful has come of this. I also have that the derivative has a max and min (based on the compact domain and the inequality condition). What am I missing here to get that $f=0$?

REPLY [3 votes]: Denote by $M$ the quantity $\max_{t\in [0,1]}|f(t)|$. Using the mean value theorem we get for $x\in (0,1)$, 
$$|f(x)|\leqslant x M.$$
We can prove by induction that $|f(x)|\leqslant x^nM$ for each $x\in (0,1)$. Indeed, it is true for $n=1$, and if it is true for $n$, then by the mean value theorem, $$f(x)=f'(c)x $$
for some $c\in (0,x)$, hence $|f(x)|\leqslant x|f'(c)|\leqslant x|f(c)|$. Using the induction hypothesis, we get $|f(x)|\leqslant xc^nM\leqslant x^{n+1}M$.
Since $x^n\to 0$ as $n$ goes to infinity, we get the wanted conclusion.