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Let $f:[0,1] \to R$ be differentiable. $f(0) = 0$ and $|f'(x)| \leq |f(x)|, \forall x \in (0,1).$

Prove that f is the $0$ function.

I can see why it's true. I can think of various counterexamples where this will not be true if $f$ is not the $0$ function. However, I'm having trouble putting together a good proof for this. Can anyone get me started on it? I think if I can get going in the right direction I'll be good to go.

Nolan P
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  • The mean value theorem should be the tool of your choice here. – Vercassivelaunos Dec 06 '20 at 22:10
  • Also: https://math.stackexchange.com/q/509876/42969, https://math.stackexchange.com/q/3423085/42969 – all found with Approach0 – Martin R Dec 06 '20 at 22:18
  • You could also use the fact that $f$ can be written as the integral of $f'$ then do some approximations to conclude that actually the two are equal in modulus the use the initial condition that you have. After this you'd have to use one more fact that is given for free if you have some ode theory under your belt, otherwise it wouldn't be too hard to prove wothout any ode theory. – THIG Dec 06 '20 at 22:26
  • @MilosTasic: You can't, because it isn't assumed to be continuosly differentiable. There are differentiable functions whose derivatives are not integrable. – Vercassivelaunos Dec 06 '20 at 22:56
  • @Vercassivelaunos yes you are correct! I suspect this could be avoided by maybe using the Henstock-Kurzweil integral but I suspect this would be beyond the context of the question! Thanks for pointing this out! – THIG Dec 06 '20 at 23:12

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