Showing the function $f(x)=0$ on $[a,b]$ Suppose for real numbers $a<b$ one has a function with continuous derivative 
$$f:[a,b]\to \mathbb{R}$$
such that $f(a)=0$ and there exists a real number $C$ with 
$$|f'(x)|\leq C|f(x)|\:\:\:\text{for all}\:\:x\in [a,b].$$
Show that $f(x)=0$ for all $x\in[a,b].$
Given $\epsilon >0$ there exists a $\delta >0$ such that for any $x\in B_{\delta}(a)$ we have $|f(x)|<\epsilon $, that is $f(x)=0$ in that neighborhood.  Note that $f$ is continuous on a compact set.
If $f$ is nonzero, there exists $x\in[a,b]$ such that $f(x)\neq 0$. Let $w=\inf\{x:f(x)\neq 0\}.$
 Hence for any  $x<w$, we have $f(x)=0. $
Now considering the following, we get $f(w)=0$ which is a contradiction. 
$$|\frac{f(w)-f(a)}{w-a}|=|f'(\xi)|\leq C|f(\xi)|, \:\:\text{where}\:\:a<\xi<w.$$ 
Is my my argument valid? I also didn't use the continuity of the derivative. Thank you!
 A: Some concerns about your attempt: 


*

*the $\varepsilon$ is not used in the sequel of the proof.

*It is not clear that $w\gt a$ and it has to be justified.


Here is a sketch of solution (the details have to be filled to not spoil the exercise).
Using the assumption $f(a)=0$ and the fundamental theorem of calculus, we have 
$$\tag{*}    f(x)=\int_a^x f'(t)  \mathrm dt  $$
hence by the triangle inequality and $ \left\lvert f'(t)\right\rvert\leqslant C \left\lvert f(t)\right\rvert$, we get
$$\tag{**}   \left\lvert f(x)\right\rvert \leqslant C\int_a^x \left\lvert f(t)\right\rvert\mathrm dt\leqslant C(x-a)\sup_{s\in (a,b)} \left\lvert f(s)\right\rvert.$$
Now, going back to $(*)$ we have 
$$\left\lvert f(x)\right\rvert \leqslant \int_a^x\left\lvert f'(t)\right\rvert\mathrm dt\leqslant C\int_a^x\left\lvert f(t)\right\rvert\mathrm dt       $$
and using (**) with $t$ instead of $x$, we get that 
$$\left\lvert f(x)\right\rvert\leqslant\frac 12  C^2(x-a)^2\sup_{s\in (a,b)} \left\lvert f(s)\right\rvert.$$
This suggests that for any $n$, there exists $c_n$ such that for any $x\in (a,b)$, $ \left\lvert f(x)\right\rvert  \leqslant c_n\left(x-a\right)^n$, where $c_n$ has to be determined.   We will find that $c_n=C^n/n! \sup_{s\in (a,b)} \left\lvert f(s)\right\rvert$ does the job.
