# Discontinuous derivative.

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 at 18:36
He probably wants a something discontinuous almost everywhere. – Git Gud Feb 1 at 18:38
– Deven Ware Feb 1 at 18:40

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

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 at 21:14

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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