There are a lot of well-known examples for "continuous at point but not continuous on a neighbourhood", "differentiable at a point but not continous on a neighbourhood" (of course then not differentiable or not continuously differentiable).

Is it correct that for a function having a continuous derivative at a point, it must have a continuous derivative on a neighborhood around that point? I can't think of any counter-examples.


No. Considering $n=1$. The set of points of continuity of the derivative of a differentiable function must be nonempty, but the set of points of discontinuity can be dense. So if you take any function that is differentiable and whose derivative is discontinuous on a dense set, any of the points of continuity of the derivative will provide a counterexample. For much more info and references, see this answer of Dave Renfro: https://math.stackexchange.com/a/112133/

For a concrete example

$$\sum\limits_{n=1}^\infty \frac1{n^2} \left(x-\frac1n\right)^2\sin\left(\frac1{x-\frac1n}\right)$$ should be a counterexample at $0$. See also Show that function is differentiable but its derivative is discontinuous.

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  • $\begingroup$ +1 Very nice example! The linked answer even suggests functions with discontinuity set of positive measure. I cannot even think about such a function. $\endgroup$ – M. Winter Sep 29 '17 at 15:20
  • $\begingroup$ Yes, pretty nice. But now can continuous derivative at a point implies just differentiable or continuity on a neighborhood? because this function is differentiable on a neighborhood. $\endgroup$ – Upc Sep 29 '17 at 15:25
  • $\begingroup$ @Xavier: If the function is not differentiable in a neighborhood of the point, you would have to clarify what you mean by continuity of the derivative at that point. $\endgroup$ – Jonas Meyer Sep 29 '17 at 15:47
  • $\begingroup$ @M. Winter: Thanks; although I was thinking roughly along those lines originally (make the standard example happen at each point $1/n$), the particular "nice" example ultimately was essentially copied from the linked question by changing the enumeration of the rationals to $1/n$. $\endgroup$ – Jonas Meyer Sep 29 '17 at 15:49
  • $\begingroup$ @XavierYang If a function is continuously differentiable at a point, then it is differentiable in a neighborhood of this point. But this derivative does not have to be continuous. $\endgroup$ – M. Winter Sep 29 '17 at 15:59

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