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I was trying to build an example of a function that is differentiable at $0$, and around $0$. But the derivative is not continuous at $0$

A family of functions that work is: (thank you Andrew D. Hwang for the general form)

$$ f(x) = \left\{ \begin{array}{ll} x^{1+\epsilon}\psi(x^{-\alpha})) & \mbox{if } x\ne0 \\ 0 & \mbox{if x=0} \end{array} \right. $$

With $\psi$ a periodic and bounded function (or a modified trig function) and $\alpha>0,\epsilon>0$

Is there an example that does not belong to this family of functions? (I have found such examples, but I am not satisfied with them because of how I built them (they are not deeply different), so I'm still interested to get ideas!)

Albert Beton
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    @SimpleArt, have a look at this: http://math.stackexchange.com/questions/292275/discontinuous-derivative – Albert Beton Oct 08 '16 at 14:55
  • Interesting question! – Simply Beautiful Art Oct 08 '16 at 14:59
  • @AlbertBeton: When you write "not based on it", could you perhaps say in more detail what you mean? Do you mean something like you don't want examples of the type $f(x) = x^{1 + \varepsilon} \psi(x^{-\alpha})$ with $\psi$ (quasi-)periodic and $\varepsilon$, $\alpha$ positive reals, or anything obtained from such a function by adding a smooth function? – Andrew D. Hwang Oct 09 '16 at 00:34
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    @AndrewD.Hwang I was looking for a function that does not seem to be based on the example I gave. It is a vague statement, but I think that the family of functions you gave basically represents all the function that would be based on this example. So thank you for that!

    Do you know if there is another family of functions that would be differentiable at $0$ with derivatives not continuous at $0$?

     – Albert Beton Oct 09 '16 at 03:48

2 Answers2

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Suppose $f$ is differentiable on an open interval about $0$, and that $f'$ is discontinuous at $0$ (but continuous elsewhere, in the interest of delimiting the structure of "the simplest examples").

Consider the "lower" and "upper" limits of $f'$ at $0$: $$ L_{-} = \lim_{\delta \to 0^{+}} \inf_{0 < |x| < \delta} f'(x),\qquad L_{+} = \lim_{\delta \to 0^{+}} \sup_{0 < |x| < \delta} f'(x). $$ By Darboux's theorem, $\lim(f', 0)$ does not exist, so $L_{-} < L_{+}$ (strict inequality), and the interval $(L_{-}, L_{+})$ is "hit by" $f'$ infinitely many times in each neighborhood of $0$.

Qualitatively, $f'$ oscillates infinitely many times (between $L_{-}$ and $L_{+}$) in every neighborhood of $0$.

This doesn't mean that every such $f$ has the form $f(x) = x^{1 + \varepsilon} \psi(x^{-\alpha})$ with $\psi$ periodic, but does indicate why common counterexamples have this form.

Andrew D. Hwang
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  • Yes, indeed while I was trying to build an example I had to deal with the theorem of Darboux. As you indicate, there is not much liberty in terms of what could happen to prevent the derivative to be continuous.

    Thank you for your detailed answer. I think the only way to build such a function would require a periodic function. I wonder if one can prove it though? I will edit the question accordingly.

     – Albert Beton Oct 09 '16 at 14:24
  • For what it's worth, a correct statement of the oscillation condition is, "For every closed, bounded subinterval $K \subset (L_{-}, L_{+})$ and every $\delta > 0$ such that $f'$ is defined on $(-\delta, \delta)$, we have $K \subset f'(-\delta, \delta)$." It's the oscillations of $f'$ that matter, of course, not periodicity of $\psi$. Periodicity is just an easy way to guarantee oscillations. :) – Andrew D. Hwang Oct 09 '16 at 14:38
  • True! Although, I'd be curious to see an example without a periodic or a modified trig function. – Albert Beton Oct 09 '16 at 15:55
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Actually your "something" is discontinuous at each $1/n.$ So there's no way that $f'(1/n)$ can even exist for $n=1,2,\dots$ You are of course interested in functions that are differentiable in a full neighborhood of $0$ whose derivatives aren't continuous at $0.$ So this is not a candidate. I think your respect for $x^2\sin(1/x)$ may have just moved up a bit.

zhw.
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  • Thank you, that explains the IVT issue.

    Well,l I was interested in another example, and yes it would had been nice if the function was differentiable around 0, but wouldn't it be surprising if the only reason why $D^1\ne C^1$ is because of a single function?

     – Albert Beton Oct 08 '16 at 23:21
  • Ok, what was the other example? Also note $x^3\sin(1/x^2)$ is another example. – zhw. Oct 08 '16 at 23:23
  • I wanted an example that would not use trig functions. But it might be impossible, I do not know. – Albert Beton Oct 08 '16 at 23:25