Given $f \in C^\infty \big( [0 , 1 ] \big)$ with $f(0)=0$ and a $0$-neighbourhood $[0,\varepsilon)$, such that $f(x)>0$ for $x\in(0,\varepsilon)$.
Is it true, that there is a $\delta >0$, such that $f_{\vert [0,\delta)}$ is monotoniously increasing? If yes, how can we proof this?
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Targon
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1I think you mean $f(x) > 0$ for $x \in (0,\epsilon)$. – Héhéhé Jul 25 '19 at 09:58
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Marginally related: Smooth function with infinite oscillation – Dave L. Renfro Jul 25 '19 at 10:14
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1Possibly $f(x) = \exp\left(-\frac{1}{x^2}\right)\left[2 - \sin \left(\frac{1}{x}\right)\right]$ with $f(0) = 0.$ – Dave L. Renfro Jul 25 '19 at 11:06
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The function:
$$f(x) = \exp(-\frac{1}{x})(\sin(\frac{1}{x}))^{2} + (\exp(-\frac{1}{x}))^{2}$$
is a possible counterexample.
(I was thinking about functions of type $x^{n} \sin(\frac{1}{x})$ which have finitely many good derivatives first and went from there)
Thanks @Héhéhé for spotting multiple shortcomings in this answer
Radost Waszkiewicz
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1Your new function seems to vanishes for all $x=1/(n \pi)$ where $n$ is a positive integer. – Héhéhé Jul 25 '19 at 10:01
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@Héhéhé vaild point. Need to add something super small monotonic and that'll be all then. I'll make yet another edit. – Radost Waszkiewicz Jul 25 '19 at 10:07
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great counterexample, thanks you two. It works, because according to Wolfram Alpha the derivative is $$\frac{e^{-2/x} \big(e^{1/x} \sin^2(1/x) - 2 e^{1/x} \sin(1/x) \cos(1/x) + 2\big)}{x^2}$$ and this function has infinitely many zeroes with changes of the sign in every neighbourhood of zero. – Targon Jul 25 '19 at 12:50