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Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.

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$f(x)=|x|$ at $x=0$? Not sure what you mean by "very". – Ron Gordon Feb 1 '13 at 18:36
He probably wants a something discontinuous almost everywhere. – Git Gud Feb 1 '13 at 18:38
Related:… – Deven Ware Feb 1 '13 at 18:40
@RonGordon: could you elaborate what you meant by $f(x)=|x|$ at $x=0$? Sine it is continuous but not differentiable I don't quite see how it fits in this context. Thank you. – Leo Jun 19 at 21:52

4 Answers 4

Haskell's answer does a great job of outlining conditions that a derivative $f'$ must satisfy, which then limits us in our search for an example. From there we see the key question: can we provide a concrete example of an everywhere differentiable function whose derivative is discontinuous on a dense, full-measure set of $\mathbb R$? Here's a closer look at the Volterra-type functions referred to in Haskell's answer, together with a little indication as to how it might be extended.

Basic example

The basic example of a differentiable function with discontinuous derivative is $$ f(x) = \begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \\ 0 & \mbox{if } x=0. \end{cases} $$ The differentiation rules show that this function is differentiable away from the origin and the difference quotient can be used to show that it is differentiable at the origin with value $f'(0)=0$. A graph is illuminating as well as it shows how $\pm x^2$ forms an envelope for the function forcing differentiablity.

enter image description here

The derivative of $f$ is $$ f'(x) = \begin{cases} 2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)&\mbox{if } x \neq 0 \\ 0 & \mbox{if } x=0, \end{cases} $$ which is discontinuous at $x=0$. It's graph looks something like so

enter image description here

Two points

The next step is to modify this example to obtain a function that is everywhere differentiable with a derivative that is continuous on all of $\mathbb R$, except for two points. To this end, consider $$ f(x) = \begin{cases} x^2 (1-x)^2 \sin \left(\frac{1}{\pi x (1-x)}\right)&\mbox{if } 0<x<1 \\ 0 & \mbox{else}. \end{cases} $$ The graph of $f$ and it's derivative look like so.

enter image description here

A cantor set of discontinuties

Now that we have a way to construct a differentiable function whose derivative is discontinuous exactly at the endpoints of an interval, it should be clear how to construct a differentiable function whose derivative is discontinous on a Cantor set constructed in the interval. For $n\in\mathbb N$ and $m=1,2,\ldots,2^n$, let $I_{m,n}$ denote one of the $2^n$ intervals removed during the $n^{th}$ stage of construction of the Cantor set. Then let $f_{m,n}$ be scaled to have support $I_{m,n}$ and to have maximum value $4^{-n}$. The function $$F(x) = \sum_{n=0}^{\infty} \sum_{m=1}^{2^n} f_{m,n}(x)$$ will be everywhere differentiable but it's derivative will be discontinuous on the given Cantor set. Assuming we do this with Cantors standard ternary set, we get a picture that looks something like so:

enter image description here

Of course, there's really a sequence of functions here and care needs to be taken to show that the limit is truly differentiable. Let $$F_N(x) = \sum_{n=1}^{N} \sum_{m=1}^{2^n} f_{m,n}(x).$$ The standard theorem then states that, as long as $F_N$ converges and $F_N'$ converges uniformly, then the limit of $F_N(x)$ will be differentiable. This is guaranteed by the choice of $4^{-n}$ as the max for $f_{m,n}$.

Increasing the measure

Again, the last example refers to the standard Cantor ternary set but there's no reason this can't be done with any Cantor set. In particular, it can be done with a so-called fat Cantor set, which can have positive measure arbitrarily close to the measure of the interval containing it. We immediately produce an everywhere differentiable function whose derivative is discontinuous on a nowhere dense set of positive measure. (Of course, care must again be taken to scale the heights of the functions go to zero quickly enough to guarantee differentiability.)

Finally, we can fill the holes of the removed intervals with more Cantor sets (and their corresponding functions) in such a way that the union of all of them is of full measure. This allows us to construct an everywhere differentiable function with derivative that is discontinuous on the union of those Cantor sets, which is a set of full measure.

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It's fairly clear that the derivative will be discontinuous at the removed endpoints. How do you ensure it is actually discontinuous on the whole Cantor set? – dfeuer Oct 14 '13 at 22:46
@dfeuer Those endpoints actually are elements of the cantor set. For example $2/3$ is in every intersection of cantor pieces. So it suffices to show for all $N$ that the function has a discontinuous derivative at those such end points if and only it is discontinuous at all $c$ in the cantor set. – MadcowD Oct 27 at 2:20

The Weierstrass function is continuous, but it's derivative is so "discontinuous" that it doesn't exist anywhere. It is not "discontinuous," but it simply doesn't exist. I'm not sure what you mean by "very discontinuous." enter image description here

This function can be defined by $$f(x) = \sum_{n=0}^\infty a^n\cos(b^n\pi x)~~~~~a\in(0,1), b\in\Bbb{Z}^+, ab>\frac{3\pi}{2}$$

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This is my favorite counterexample in real analysis, but I’m not sure if the OP wants a non-existent derivative or one that exists but is discontinuous on a non-trivial set of points. – Haskell Curry Feb 1 '13 at 21:14

I guess that you are looking for a continuous function $ f: \mathbb{R} \to \mathbb{R} $ such that $ f $ is differentiable everywhere but $ f' $ is ‘as discontinuous as possible’.

We have the following theorem in real analysis.

Theorem 1 If $ f: \mathbb{R} \to \mathbb{R} $ is differentiable everywhere, then the set of points in $ \mathbb{R} $ where $ f' $ is continuous is non-empty. More precisely, the set of all such points is a dense $ G_{\delta} $-subset of $ \mathbb{R} $.

Note: A $ G_{\delta} $-subset of $ \mathbb{R} $ is just the intersection of a countable collection of open subsets of $ \mathbb{R} $.

The proof of Theorem 1 is an application of the Baire Category Theorem, and it can be found in Munkres’ Topology. By this theorem, it is therefore impossible to find an $ f: \mathbb{R} \to \mathbb{R} $ whose derivative exists but is discontinuous everywhere.

There is another theorem that provides a necessary and sufficient condition for a function $ g: \mathbb{R} \to \mathbb{R} $ to have an antiderivative.

Theorem 2 A function $ g: \mathbb{R} \to \mathbb{R} $ has an antiderivative if and only if its set of discontinuities is a meagre $ F_{\sigma} $-subset of $ \mathbb{R} $.

Note: An $ F_{\sigma} $-subset of $ \mathbb{R} $ is just the union of a countable collection of closed subsets of $ \mathbb{R} $.

Let me end off with a non-trivial example to add to yours. Volterra’s Function is differentiable everywhere, but its derivative is discontinuous on a set of positive measure, not just at a single point.

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consider $f(x) = \sum_{n=0}^{\infty} \frac{1}{2^n}\cos(3^nx)$ this function is continuous everywhere but the derivative exists nowhere.

this was discovered in 1872 by Karl Weierstrass

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