TITLE: Prove that $f(x)\equiv0$ on $\left[0,1\right]$ if $f(0)=0$ and $|f'(x)|\le|f(x)|$
QUESTION [9 upvotes]: Let $f(x)$ differentiable on $\left[0,1\right]$ such that $f(0) = 0$.  Also, assuming that $\forall x\in \left[0,1\right]:\left|f'(x)\right| \le \left|f(x)\right|$. Prove that $f(x)\equiv 0$  
What I did so far:
Using LMVT:
$$\frac{f(x)-f(0)}{x-0} = \frac{f(x)}{x} = f'(c)$$ where $c\in \left(0,x\right)$
$$\left|f('c)\right| = \left|\frac{f(x)}{x}\right|.$$  
Since, $\left|x\right| \in (0,1)$
$$\left|\frac{f(x)}{x}\right| \ge \left|f(x)\right| \ge \left|f'(x)\right|$$
So, $$\left|f('c)\right| \ge \left|f'(x)\right|$$
But I'm not sure how is it helping me. What am I missing?

REPLY [2 votes]: This is similar to an older question 
Should $f(x) \equiv 0$?.
Consider $g(x)=\mathrm e^{-x}f(x)$. Then $g'(x)=\mathrm e^{-x}\left(f'(x)-f(x)\right)$. Consider a fixed $x\in[0,1]$. If $f(x)\geq0$ then $g(x)\geq0$ and since $\lvert f(x)\rvert>\lvert f'(x)\rvert$, we have $g'(x)\leq0$. Conversely, if $f(x)\leq0$ then $g(x)\leq0$ and $g'(x)\geq0$. So we have, for all $x\in[0,1], g(x)g'(x)\leq0$. And consequently $h(x)=\frac12g(x)^2$ is decreasing on $[0,1]$. As $h(0)=0$ and $h(x)\geq0$ we conclude that $h(x)=0$ for all $x\in[0,1]$ and thus $f(x)=0$ for all $x\in[0,1]$.