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Does there exists an $f:\mathbb{R}\rightarrow \mathbb{R}$ differentiable everywhere with $f'$ discontinuous at some point?

Kal S.
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    Cantor-Volterra's functionis an example of differentiable function whose derivative has discontinuity set of positive measure. – Sangchul Lee May 02 '14 at 03:22

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Yes. Do you know an example of a continuous function that is not differentiable at some point? (Hint: think of a corner) If you integrate it....

Ross Millikan
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  • OK got it, thank you. I was thinking it from another point of view, didn't try integrating a function. – Kal S. May 02 '14 at 03:22
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    The point of the exercise is that each level of differentiability brings something new in information about (restricting) the function. You can have a function that is $n$ times differentiable but not $n+1$ times. – Ross Millikan May 02 '14 at 03:27
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    Perhaps I'm missing the point but surely this does not answer the question. It seems to me that this gives a function $f$ for which $f'$ is not differentiable - but the question asked about a function $f$ for which $f'$ is not continuous. – David May 02 '14 at 03:55
  • @David: reading again, you are correct. I should have suggested integrating a step function. The first function that came to mind for my example was $f(x)=|x|$, but when integrated the derivative is continuous. – Ross Millikan May 02 '14 at 04:03
  • Sorry but I don't think a step function works either: if $f$ is the integral of a step function then $f'$ does not exist at the "step". I think my example works, would you care to check it out and let me know what you think? – David May 02 '14 at 04:08
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Yes there does exist such a function, for example $$f(x)=\cases{x^2\sin(1/x)&if $x\ne0$\cr 0&if $x=0$.\cr}$$ By normal differentiation rules we have $$f'(x)=2x\sin(1/x)-\cos(1/x)$$ if $x\ne0$, and for $x=0$ we use the definition: $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h-0}=\lim_{h\to0}h\sin(1/h)=0\ .$$ So $f$ is certainly differentiable at $0$, and in fact everywhere else too. However, if $x\to0$ then $2x\sin(1/x)\to0$ and $\cos(1/x)$ oscillates between $1$ and $-1$, so $f'(x)$ has no limit as $x\to0$, and is therefore not continuous at $x=0$.

David
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  • $\lim_{h \to 0} h \sin (1/h)=1$, but your argument works well. – Ross Millikan May 02 '14 at 04:15
  • @David Thank you, I realize that the other function suggested indeed didn't work. Also, there is a post about this particular function. I am posting it for future reference: http://math.stackexchange.com/questions/232672/show-that-the-function-gx-x2-sin-frac1x-g0-0-is-everywhere – Kal S. May 02 '14 at 04:27