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Edited:

It is known that if $f$ is differentiable then the derivative function of $f$ is not always continuous. For instance $f(x)=x^2\sin (\frac{1}{x})$ for $x\neq 0$ and $f(0)=0$ if $x=0$. Then $f^{\prime}$ is discontinue at $x=0$.

Is there any differentiable function $f$ whose the derivative of $f$ has countable points of discontinuity?

bung
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  • Well, $|x|$ is not differentiable. – Thomas Apr 24 '16 at 07:39
  • @Thomas : $|x|$ is weak differentiable, very different to $\sqrt{|x|}$ which is not differentiable. – reuns Apr 24 '16 at 07:42
  • and $|x|$ symmetrized and periodized is an example of a weak differentiable function with a countably infinite number of discontinuity points – reuns Apr 24 '16 at 07:43
  • @user1952009: Great as it is, the question is on differentiability and not weak differentiability. – Mathematician 42 Apr 24 '16 at 07:44
  • @Mathematician42 : yes but the answer is that such a function is not differentiable but is weak differentiable – reuns Apr 24 '16 at 07:46
  • Also the title of this question seems to suggest that the poster believes any continuous function is to some degree differentiable. This is not the case (which is sort of remarkable). – Mathematician 42 Apr 24 '16 at 07:47
  • @user1952009: Consider the function $f(x)=x^2\sin(\frac{1}{x})$ for all $x\in \mathbb{R}_0$ and $f(0)=0$. This is a differentiable function whose derivative is discontinuous. – Mathematician 42 Apr 24 '16 at 07:50
  • @user1952009 I know that $|x|$ is weakly differentiable. The orginal question, however was about differentiable functions for which the derivative fails to be continuous and not about weakly differentiable functions. This something different, since there are functions wich are everywhere differentiable (in the sense of the classical definition, not in the weak sense, the OP changed his question and now is using such a function) without continuous derivative. I do not see that changing to weak differentiability is desired nor necessary in this discussion. It is then a different question. – Thomas Apr 24 '16 at 08:23

2 Answers2

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As noted in the comments, $f(x) = |x|$ is not differentiable at $x = 0$ and thus whatever the derivative $f'$ as a function may be, it isn't continuous at $x = 0$ either because $f'(0)$ doesn't exist. The added complication in this case is that

$$\lim_{x \to 0} f'(x)$$ does not exist either.

It is relatively straight forward to construct a function with countably many points $\{ p_i \}$ where the derivative doesn't exist, even if all of the limits

$$\lim_{x \to p_i} f'(x)$$

exist.

For example

$$f(x) = \begin{cases} 0, & \lfloor x \rfloor \text{ is even } \\ 1, & \lfloor x \rfloor \text{ is odd } \end{cases} $$

For all $x \in \mathbb R - \mathbb Z$, the derivative exists and is identically zero. Thus the limits

$$\lim_{x \to n} f'(x) = 0 \quad \text{ for all integer } n$$

If this isn't the kind of discontinuity you are interested in, please clarify.

Simon S
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Differentiability is a local question. Therefore, what you need to know in order to answer your question in the affirmative are basically two things:

i) a differentiable function on an open interval the derivative of which fails to be continuous in exactly one interior point - you provided such a function in your question after an edit ($x^2 \sin(x))$

ii) the existence of a countable disjoint set of open balls in the domain of definition you are interested in. You can then, with the aid of i) patch such a function together, e.g. by using smooth cutoff functions with center of mass in the center of those balls and vanishing smoothly at (and in a even smaller neighbourhood of) their boundary.

in case you are looking at a function defined on $\mathbb{R}$, you can just take the intervals $(n-\frac{1}{2},n+\frac{1}{2})$ whith $n\in \mathbb{Z}$, with the discontinuities located at $n\in \mathbb{Z}$. Slightly more interisting is the case that your function should live on a compact interval (e.g. $[0,1]$), but also in this case it is known (and easy to see, using an enumeration of the rational numbers and inductively choosing balls of radius $\frac{1}{2^n}$, say), that ii) is true.

Thomas
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  • Can you give me an example of differentiable function $f$ on $\mathbb{R}$ whose $f^{\prime}$ has countable points of discontinuity? – bung Apr 24 '16 at 11:37
  • @bung I told you how you can construct one, it should be quite easy though a bit tedious to write one down explicitly on your own following the instructions. – Thomas Apr 24 '16 at 13:03