Let $f$ be differentiable on $[a,b]$, $f(a)=0$ and there is $c\geq 0$ such that $|f'(x)|\leq c|f(x)|$ then $f(x)=0$ Prove that if  $f$ be differentiable on $[a,b]$, $f(a)=0$ and there is a constant  $c\geq 0$ such that $|f'(x)|\leq c|f(x)|$ for all $x\in [a, b]$,  then $f(x)=0$.
Any piece of advice would be much appreciated.
 A: It would be easier if you knew $f(x) \ge 0$; perhaps the following works?
Look at an interval $[a,y]$ with $y \in (a,b]$ where $f$ has a constant sign, suppose for example (WLOG) that $f(x) \ge 0$ on $[a,y]$ and consider the function $g(x)=f(x)e^{-cx}$, then:
$$g'(x) = \left(f'(x)-cf(x)\right)e^{cx} \le 0$$
This means $g$ is non-increasing on $[a,y]$ but $g(a)=0$ and $g(x)\ge 0$ on $[a,y]$ so...
A: For $c=0$ the result is obvious. So, assume that $c>0$. 
Let $x_0=a+1/2c$, $M_0=sup|f(x|, M_1=sup|f'(x)|$ on $[a, x_0]$. 
Now, according to mean-value theorem, for $x\in (a, x_0)$:
$$|f(x)|\leq M_1(x_0-a)\leq cM_0(x_0-a)=M_0/2$$
which means $M_0\leq M_0/2$, and so $M_0=0$. 

Hence, $f(x)=0$ on $[a, x_0]$. 

Now, repeat the process replacing $a$ with $x_0$. Since $b-a$ is a finite, after a finite number of iterates, the result is at hand.
A: choose some $x_0∈ (a, b]$ such that $A(x_0−a) < 1$. If $M_0 = 0$, then clearly $f = 0$, and we can proceed. If $M_0 > 0$, note that
 $f′(x) ≤ A |f(x)| ≤ AM_0$
for $x ∈ [a, b]$, so $M_1 ≤ AM_0$. Now, denote $δ = 1−A(x_0−a)$, then for $x ∈ [a, x_0]$, $|f(x)| =f'(c)(x − a) ≤ M-1(x_0 − a) ≤ A(x_0 − a)M_0 = M_0 − δM$
where $c ∈ (a, x)$ exists because of the mean value theorem. It follows that $M_0 − δM_0$ is a new upper bound of $|f(x)| for x ∈ [a, x_0]$, contradicting the definition of $M_0$. So $M_0 > 0$ is impossible, and we conclude that $f = 0$ on $[a, x_0]$. Next, choose $x_k= (k + 1)x_0 − ka$ for $k = 0, 1, 2, . . . , n− 1$, where $n ≥ 1$ satisfying $x_n= b$ and $x_n − x_{n-1} ≤ x_{0}− a$. By using the same argument, $f = 0$ on $[x_{k-1}, x_k]$, for $k = 1, 2, . . . , n$. Hence $f = 0$ on $[a, b]$
