Prove that $f(x) = 0$ for all $x \in [a,b]$, where $f$ is differentiable on $[a,b],~f(a) = 0$ and $|f'(x)| \leq A|f(x)|$ on $[a,b]$ for some $A\geq0$ Q. Suppose $f$ is differentiable on $[a,b]$, $f(a) = 0$ and there is real number $A$ such that $|f'(x)| \leq A|f(x)|$ on $[a,b]$. Prove that $f(x) = 0$ for all $x \in [a,b]$.
My attempt: if $A=0$, it is clear that $f(x)=f(a)=0$ for all $x \in [a,b]$.
If $A>0$, take $x_0=a+\frac{1}{nA},~n \neq 0$, then, from the mean value theorem and $M_1 \leq AM_0$, it is clear that $$|f(x)| \leq  M_1 (x - a)~\&~M_1(x-a) \leq A M_0 (x_0 - a)$$
for any $x \in [a,x_0]$, where $M_1=\sup_{[a,x_0]}|f'(x)|$ and $M_0=\sup_{[a,x_0]}|f(x)|$, this gives $$|f(x)| \leq A M_0 (x_0 - a)=\frac{M_0}{n},$$
for any $x \in [a,x_0]$. Hence $$M_0=\sup_{[a,x_0]}|f(x)| \leq \frac{M_0}{n},$$
Now,
(i) How can I ensure that $n>1$ to obtain $M_0=0$ ?
(ii) How the idea is extended to any $x_0 \in (a,b]$ ?
 A: You have the right idea. I think the best way to proceed is to establish that if $x\in [a,b]$ such that $f(x)=0$, then $f(x)=0$ on $\left[ x,x+\frac 1{2A}\right]\cap [a, b]$ (in your solution, we are free to fix $n$ at any value, though the proof only works if $n>1$. I'll pick $n=2$ for convenience). The technique you used looks correct to me, all you need to do is replace $a$ with $x$.
From here, we can use an inductive argument. You've shown that the claim holds for $x=a$. Applying the above once tells us that $f=0$ on $\left[a,a+\frac{1}{2A}\right]$. Applying it again gives us that $f=0$ on $\left[a,a+\frac{2}{2A}\right]$, and so on and so forth. Since $b$ is finite, we'll eventually "cover" all of $[a,b]$.
A: Another version of the above:
Suppose $K = \max_{x \in [a,b]} |f(x)|$ and $y \in [a,b]$ such that $f(y) = 0$. Consider $x\in [y,b]$, then clearly
$|f(x)-f(y)| \le K$.
I claim that $|f(x)-f(y)| \le K (A(x-y))^n$ for $n=0,1,...$. It is true for $n=0$, suppose it is true for $n$, then
$|f(x)-f(y)| \le |f'(\xi)| (x-y)$ for some $\xi \in [y,x]$ and so
$|f(x)-f(y)| \le AK (A(x-y))^n (x-y) = K (A(x-y))^{n+1}$.
In particular, if we choose $\delta >0$ such that $A\delta <1$ then we see that for $x \in [y,y+\delta]$ that $f(x) = f(y) = 0$.
The same argument and $\delta$ apply starting at $y+\delta$, hence it follows by induction that $f(x)=0$ for all $ x\in [a,b]$.
This is reminiscent of the Picard iteration.
