Prove that $f(x)\equiv0$ on $\left[0,1\right]$ if $f(0)=0$ and $|f'(x)|\le|f(x)|$ 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?  

Assume $f$ is not identically zero on $[0,1]$. Since $f$ is continuous on $[0,1]$, there is a maximum point $b\in[0,1]$ such that $|f(b)|\geq |f(x)|$ for all $x\in [0,1]$. If possible, we will choose $b<1$. In other words, either $b<1$, or $b=1$ and $|f(x)|<|f(b)|$ for all $x\in[0,b)$. Without loss of generality, we may assume that $f(b)>0$. (Otherwise consider the function $-f(x)$.)
Use the mean value theorem on the interval $[0,b]$. This yields a point $c\in (0,b)$ such that $f'(c)={f(b)-f(0)\over b-0}={f(b)\over b}$. There are now two cases:
(i) If $b<1$ then $f'(c)={f(b)\over b}>f(b)\geq |f(c)|$, which contradicts $|f'(x)|\leq |f(c)|$.
(ii) If $b=1$ then $f'(c)=f(b)>|f(c)|$, again contradicting $|f'(x)|\leq |f(c)|$.
