differentiable function with $f=0$ Let $f:[a,b]\rightarrow \mathbb{R}$ a differentiable function for which there exists $C>0$ such that $|f'(x)|\leq C|f(x)|$. Show that if $f(a)=0$, then $f=0$.
I tried to use Rolle's Theorem.
 A: Since $f$ is continuous on $[a, b]$ then it is bounded. This impiles that $\sup_{x\in [a, b]}|f'(x)|\leqslant C \sup_{x\in [a, b]}|f(x)|:=M<\infty$. From the assumption we have for each $x\in [a, b]$,
$$|f'(x)|\leqslant C \Big|\int_a^x f'(t)dt\Big|\leqslant C\int_a^x|f'(t)|dt\leqslant C^{n+1} \int_a^x \int_a^{t_1}\cdots \int_a^{t_n} |f'(t_{n+1})|dt_{n+1} \leqslant C^{n+1} M \frac{x^{n+1}}{(n+1)!}$$
for all $n$.  Let $n\to \infty$, the RHS tends to $0$ and this implies $f'(x)=0$ on $[a, b]$. Combine with $f(a)=0$ we obtain $f(x)\equiv 0$ on $[a, b]$.
Another way: By using Rolle's theorem,
$\bullet$ If $C<\frac{1}{b-a}$, then
$$f'(x)\leqslant C|f(x)|=C|f(x)-f(a)|=C|f'(\xi_1)||x-a|\leqslant C|b-a||f'(\xi_1)|\leqslant \cdots \leqslant (C(b-a))^n |f'(\xi_n)| \leqslant (C(b-a))^n M \to 0 $$
as $n\to \infty$ as $C(b-a)<1$, where $\xi_k \in (a, \xi_{k-1})$, $\xi_0=x$. Hence, in this case $f'(x)=0$ for every $x\in [a, b]$ and then $f(x)=0$ on $[a, b]$.
$\bullet$ If $C\geqslant \frac{1}{b-a}$, or equivalently $a+ \frac{1}{C}\leqslant b$. We devide the inteval $[a, b]$ into $k$ part as follows: $[a, a+\frac{1}{C}], [a+\frac{1}{C}, a+ \frac{2}{C}],\cdots, [a+\frac{k}{C}, b]$, where $k=\max\{m : a+ \frac{m}{C}\leqslant b\}$. We will prove that $f(x)=0$ on each such part. Indeed, for $x\in (a, a+ \frac{1}{C})$, by the assumption
$$|f'(x)|\leqslant C|f(x)|=C|f(x)-f(a)|=C|f'(\xi_1)||x-a|\leqslant \cdots \leqslant (C(x-a))^n M.$$ 
Since $C(x-a) <1$ in this case, we have $f'(x)=0$ after taking $n\to \infty$. Thus, $f'(x)=0$ for all $x\in (a, a+\frac{1}{C})$. By the continuity of $f$, we get $f(x)=0$ on $[a, a+\frac{1}{C}]$. Similarly for remaining intervals, we get $f(x)=0$ on $[a, b]$.
