# Characterizing discontinuous derivatives

Apparently the set of discontinuity of derivatives is weird in its own sense. Following are the examples that I know so far:

$1.$ $$g(x)=\left\{ \begin{array}{ll} x^2 \sin(\frac{1}{x}) & x \in (0,1] \\ 0 & x=0 \end{array}\right.$$ $g'$ is discontinuous at $x=0$.

$2.$ The Volterra function defined on the ternary Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Cantor set ,that is on a nowhere dense set of measure zero.

$3.$ The Volterra function defined on the Fat-Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Fat-Cantor set ,that is on a set of positive measure, but not full measure.

$4.$ I am yet to find a derivative which is discontinuous on a set of full measure.

Questions:

1.What are some examples of functions whose derivative is discontinuous on a dense set of zero measure , say on the rationals?

2.What are some examples of functions whose derivative is discontinuous on a dense set of positive measure , say on the irrationals?

Update: One can find a function which is differentiable everywhere but whose derivative is discontinuous on a dense set of zero measure here.

• The Volterra function is not defined just on the Cantor set, but the whole interval. It equals $0$ on the Cantor set. I know you know this. But perhaps "defined relative to the Cantor set" might be more accurate. – zhw. Feb 24 '18 at 17:32
• @zhw Yeah, but I think its pretty obvious though. :P – yasir Feb 24 '18 at 17:44
• "Some good discussion about this can be found here and here." The first link you gave gives several references to the fact that any $F_{\sigma}$ first category set (equivalently, any countable union of closed nowhere dense sets) is the discontinuity set of some derivative. Those references that give a proof of this result do so in a constructive way, in the sense that given any countable union of closed nowhere dense sets (note that $\mathbb Q$ is a countable union of singleton sets), a differentiable function is constructed whose derivative is discontinuous exactly on that countable union. – Dave L. Renfro Feb 24 '18 at 21:08

There is no everywhere differentiable function $f$ on $[0,1]$ such that $f'$ is discontinuous at each irrational there. That's because $f',$ being the everywhere pointwise limit of continuous functions, is continuous on a dense $G_\delta$ subset of $[0,1].$ This is a result of Baire. Thus $f'$ can't be continuous only on a subset of the rationals, a set of the first category.

But there is a differentiable function whose derivative is discontinuous on a set of full measure.

Proof: For every Cantor set $K\subset [0,1]$ there is a "Volterra function" $f$ relative to $K,$ which for the purpose at hand means a differentiable function $f$ on $[0,1]$ such that i)$|f|\le 1$ on $[0,1],$ ii) $|f'|\le 1$ on $[0,1],$ iii) $f'$ is continuous on $[0,1]\setminus K,$ iv) $f'$ is discontinuous at each point of $K.$

Now we can choose disjoint Cantor sets $K_n \subset [0,1]$ such that $\sum_n m(K_n) = 1.$ For each $n$ we choose a Volterra function $f_n$ as above. Then define

$$F=\sum_{n=1}^{\infty} \frac{f_n}{2^n}.$$

$F$ is well defined by this series, and is differentiable on $[0,1].$ That's because each summand above is differentiable there, and the sum of derivatives converges uniformly on $[0,1].$ So we have

$$F'(x) = \sum_{n=1}^{\infty} \frac{f_n'(x)}{2^n}\,\, \text { for each } x\in [0,1].$$

Let $x_0\in \cup K_n.$ Then $x_0$ is in some $K_{n_0}.$ We can write

$$F'(x) = \frac{f_{n_0}'(x)}{2^{n_0}} + \sum_{n\ne n_0}\frac{f_n'(x)}{2^n}.$$

Now the sum on the right is continuous at $x_0,$ being the uniform limit of functions continuous at $x_0.$ But $f_{n_0}'/2^{n_0}$ is not continuous at $x_0.$ This shows $F'$ is not continuous at $x_0.$ Since $x_0$ was an aribtrary point in $\cup K_n,$ $F'$ is discontinuous on a set of full measure as desired.

• @ zhw So, if $f'$ is discontinuous on a set $D$, then $D$ can be dense only if its Lebesgue measure is zero? – yasir Feb 25 '18 at 3:04
• No, certainly not. A set of full measure is dense. – zhw. Feb 25 '18 at 7:09
• @ zhw. Is the set on which $F'$ is continuous dense, as per your claim in first paragraph? – yasir Feb 25 '18 at 9:26
• @yasir Yes, it has to be. – zhw. Feb 25 '18 at 17:14