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?  

Let $x_0 \in [0,1]$ with $f(x_0) = 0$. We will show that $f(x) = 0$ for all $x \in [x_0, x_0 + \varepsilon]$ for some $\varepsilon > 0$ (if $x_0 < 1$). This is sufficient to prove your claim because you can then consider $\gamma := \sup \{x \in [0,1] \mid f|_{[0,x]} \equiv 0\}$, show $\gamma = 1$ and deduce $f \equiv 0$.
Because $f$ is continuous and $f(x_0) = 0$, there is some $\varepsilon_1 \in (0, \frac{1}{2})$ with $|f(x)| \leq \frac{1}{2}$ for all $x \in [x_0 , x_0 + \varepsilon]$.
We will now show inductively that actually $|f(x)| \leq \left(\frac12\right)^n$ for all $x \in [x_0, x_0 + \varepsilon]$. For $n=1$ this follows by our choice of $\varepsilon$.
For the inductive step note that $|f'(x)| \leq |f(x)| \leq \left(\frac12\right)^n$ for all $x \in [x_0, x_0 + \varepsilon]$. Using the mean-value theorem, it is now easy to see that
$$|f(x)| = |f(x) - f(x_0)| = |f'(\xi)| \cdot |x - x_0| \leq \left(\frac{1}{2}\right)^n \cdot |x - x_0| \leq \left(\frac{1}{2}\right)^{n+1}$$
for $x \in [x_0, x_0 + \varepsilon]$, because $\varepsilon \leq \frac{1}{2}$.
In the limit $n\rightarrow \infty$, we get our claim.
