How to show that $f(x)=0 $ for all $x\in [0,1]?$ This is a question from a sample paper of a entrance exam. I did this type of problems using MVT but this time i dont have any hint how to solve this
The question is
 "let $ f :[0,1]- R $ is differentiable and $ f(0)=0. $ Given that there exist a $c\in (0,1)$  such that $ |f'(x)|  \leq c|f(x)|$  for all $ x \in [0,1]. $ Then prove that $ f(x) =0$ for all $x\in [0,1].$
I was thinking if i can prove this by contradiction. That is if I assume $ f(x)>0 $ or $ f(x)<0 $ then by MVT i can arrive at a contradiction. But that's where i am stuck now since $c\in (0,1) $ then $c<1$  then $|f'(x)|<|f(x)|$  but i dont know what to do after that. Next i have some idea if i can show that $f'(x) = 0$  for all $x $ then $ f $ is a constant function and since $f(0)=0 $ then the whole function is zero
. That's why i need some hints to slove this problem. 
 A: Hint: suppose the maximum absolute value of $f$ is attained at some $y\in [0,1]$ (by the extreme value theorem). Then 
\begin{equation}
\vert f(y)\vert=\bigg\vert \int_0^y f'(x)dx\bigg\vert\leq \int_0^y \vert f'(x)\vert dx\leq c\int_0^y \vert f(x)\vert dx\leq c\cdot y\cdot \vert f(y)\vert.
\end{equation}
What happens if $\vert f(y)\vert\neq 0$?
A: As I mentioned in a comment, I'm not sure what is wrong with the answer of Eeyore Ho. Here is an elaboration.
Considering instead the differentiable function $g:=f^2=|f|^2$, we note that the assumptions imply
$$ g' \le 2c g$$
One can now use Gronwall's inequality to finish, or indeed just follow the proof of Gronwall's inequality. Consider the function
$$ h(x) = g(x)e^{-2cx}$$
then
$$ h'(x) = (g'(x)-2c g(x))e^{2cx} \le 0$$
So $h$ is non-increasing. But $h(0) = 0 $ and $h\ge 0$, so $h(x)=0$ for all $x$; so $g(x)=0$ for all $x$, so $f(x)=0$ for all $x$.
PS this doesn't require $c\in(0,1)$.
A: I will use the mean value theorem. 
Let $a\in [0,1]$ we have by MVT $a_1\in [0,a]$ such that  $\frac{f(a)- f(0)}{a} = f'(a_1)\leq c f(a_1)$ thus $f(a) \leq ac f(a_1)$
Again for $a_1 \in [0,a_1]$ we have $a_2\in [0,a_1]$ such that $\frac{f(a_1)-f(0)}{a_1-0} = f'(a_2) \leq c f(a_2)$ combine the first and the second we get 
$f(a) \leq ac f(a_1) \leq aa_1c^2f(a_2)$  (note that $c\in (0,1)$ also $a,a_1 \leq 1$ )
So $f(a) \leq c^2 f(a_1)$ keep going we get 
$f(a) \leq c^n f(a_n)$ 
A: Let $g(x)=f^{2}(x),\, g(x)\geq 0,$
$$\left |g^{'}(x)\right |\leq 2cg(x)$$
Prove $g(x)=0$ for all $x\in[0,1]. $
And then you can get
$ f(x)=0$ for all $x\in[0,1]. $
