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$f$ is differentiable on $[a,b]$, $f'(x) \leq A|f(x)|$ where $A$ is a non-negative constant.

If $f(a)=0$ show $f(x)=0, \forall x\in [a,b]$

I imagine the proof uses the Mean Value Theorem but I have not been able to get it to work.

I know $|f(x)|=|f'(c)|(x-a)$ where $c \in [a,x]$, so $|f(x)| \leq A\ |f(c)|(x-a)$ where $c\leq x$ And I guess I could sort of iterate this to keep getting a smaller $c$ but I don't see why it must go all the way to zero.

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    Is this $|f'(x)|\leqslant A|f(x)|$? Then: Gronwall's lemma.2012-01-11
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    Iterating the final inequality seems like the way to go. But I doubt that the point is that you keep "getting a smaller c". On the other hand, notice that you get an extra $A(x-a)$ factor; if $x$ is enough close (how close?) to $a$, this factor is strictly smaller than $1$. On iterating this $n$ times, you will get an $(A(x-a))^n$, which approaches $0$ as $n \to \infty$. Does this help? [There are some details to fill, but the idea is hopefully clear.]2012-01-11

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