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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?

Targon
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1 Answers1

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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
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    This function does not seem to be smooth. – Héhéhé Jul 25 '19 at 09:44
  • Sry, fixed my mistake – Radost Jul 25 '19 at 09:54
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    Your new function seems to vanishes for all $x=1/(n \pi)$ where $n$ is a positive integer. – Héhéhé Jul 25 '19 at 10:01
  • @Héhéhé vaild point. Need to add something super small monotonic and that'll be all then. I'll make yet another edit. – Radost Jul 25 '19 at 10:07
  • 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