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I was just reading a bit about continuosly differentiable functions and I found that one 'standard' example of a function that differentiable but not continuously differentiable is $f(x)=x^2\sin(1/x)$ for $x\neq 0$ and $f(0)=0$. Now I have the following conjecture:

Let $f: \mathbb{R}^+_0 \to \mathbb{R}$ a continuously differentiable function. Asume that there exists $s=\inf\{ t>0: f(t)<0 \}.$ Is it true that it is always possible to find an interval $J=(s,s+\delta)$ such that for every $t\in J$,$f(t)<0$?

I cannot find a counterexample (a function that oscillates near $s$) for this, but I am stuck showing its validity. Can you please give some help?

Thanks in advance

rowcol
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    Maybe something like $f(x) =x^{10} sin (1/x)$ for $x > 0$ and $f(x)=0$ for $x\le 0$. Then $f$ is continuously differentiable, $s=0$ as you have defined it, but $f$ is both positive and negative infinitely often in any neighborhood of $0$. – User8128 Apr 04 '22 at 04:04
  • I don't understand that purpose of the requirement "Assume there exists $s=\inf{ t>0: f(t)<0 }.$" Since every nonempty set has an infimum, this requirement is simply "there exists $t>0$ such that $f(t)<0$". Maybe you mean "assume there exists $t>0$ such that $f(t)<0,$ and let $s=\inf{ t>0: f(t)<0 }$", although this doesn't work and you probably want to use $s = \inf{ t\geq 0: f(t)<0 }.$ That said, possibly the example I give in this answer (due to Dini, from 1878) will be of interest. – Dave L. Renfro Apr 04 '22 at 07:43
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    although this doesn't work --- Oops, defining $s$ by $s=\inf{ t>0: f(t)<0 }$ is OK, although $s=0$ is possible. However, I don't see any reason why we would want to require $s>0.$ – Dave L. Renfro Apr 04 '22 at 08:42
  • Just by the way! The example you called "standard" is part of a family that everyone should know well. It can be used for a multitude of counterexamples. Read this article if you can find it: Kaptanoğlu, H. Turgay, In praise of $y=x^αsin(1/x)$. Amer. Math. Monthly 108 (2001), no. 2, 144–150. – B. S. Thomson Apr 04 '22 at 16:45

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