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If function $f$ is continuously differentiable at some point, say $x=0$, and is Lipschitz in some neighborhood of $x=0$, is that true there is an open neighborhood of $x=0$ in which $f$ is continuously differentiable?

I know there is a function which is differentiable at just one point and continuous everywhere else. I also know the set of continuity of a derivative of a function is dense. I also familiar with a differentiable function which is not $C^1$ On the Cantor set.

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    $\begingroup$ What does it mean for a function to be continuously differentiable at a point? $\endgroup$ – user258700 Dec 25 '15 at 23:44
  • $\begingroup$ It means $\lim_{x\to x_0}f'(x)=f'(x_0)$. $\endgroup$ – Mehrdad Dec 26 '15 at 10:45
  • $\begingroup$ This doesn't make sense unless $f'$ exists in a neighbourhood of $x_0$. $\endgroup$ – user258700 Dec 26 '15 at 10:54
  • $\begingroup$ Yes, of course, we suppose $f'$ exists in a neighborhood of $x_0$ $\endgroup$ – Mehrdad Dec 26 '15 at 11:08
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No it is not true.

Let $$ f(x)=\max\{1-\lvert x\rvert,0\}, $$ and define $$ g(x)=\sum_{n=1}^\infty 2^{-n}f\Big(nx-\frac{1}{n}\Big). $$ Then $f$ is Lipschitz everywhere, differentiable a $x=0$, and not differentiable at $x=1/n^2$, for all $n\in\mathbb N$.

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  • $\begingroup$ Thank you for your answer, but I need the function be continuously differentiable at x=0, not just differentiable. $\endgroup$ – Mehrdad Dec 26 '15 at 10:39
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Construct any everywhere differentiable function $f$ such that $f'$ has a dense set of discontinuities. The function $f'$ has to be continuous at some point $x_0$ (since $f'$ is Baire 1) and so at that point you won't have an open neighborhood where $f'$ is continuous. For the construction see

https://math.stackexchange.com/questions/292275/discontinuous-derivative

Note that in some neigborhood of $x_0$ the function $f'$ is bounded and so the function $f$ is Lipschitz in that interval. (So, in fact, asking for $f$ to be Lipschitz in a neigborhood was redundant since continuity of the derivative at a point supplies that anyway.)

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  • $\begingroup$ Thank you for your answer. $\endgroup$ – Mehrdad Dec 26 '15 at 22:18

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