# Does there any diffentiable function $f$ such that $f'$ is discontinuous exactly on $\Bbb{Q}$ and continuous on $\Bbb{R}\setminus \Bbb{Q}$?

Does there exists any diffentiable function $$f$$ such that $$f'$$ is discontinuous exactly on $$\Bbb{Q}$$ and continuous on $$\Bbb{R}\setminus \Bbb{Q}$$ ?

Since $$\Bbb{Q}$$ is $$F_{\sigma}$$ , we can produce a function which is discontinuous only on $$\Bbb{Q}$$.

For an example we can pick Thomae's function.But Thomae's function has no primitive. Because if thomae's function $$f$$ has a primitive $$F$$ then $$F'=f$$ . Since $$F'$$ is Darboux function, image of $$F'=f$$ must contains an intervals and this is not possible as Thomae's function doesn't attain irrational values.

By choosing a particular example, we can conclude the impossibility of existence of such function.

If $$f'$$ is Darboux function and belongs to Baire class $$1$$ then $$f'$$ has a primitive $$f$$ .

Hence our goal is to create a Darboux function $$f'$$ of Baire class $$1$$ which is continuous on $$\Bbb{Q}$$ and discontinuous on $$\Bbb{R}\setminus \Bbb{Q}$$.

How to produce such function?

• Dave Renfro's comprehensive answer to How discontinuous can a derivative be? is probably of interest. Jun 15 at 1:36
• Just saw this question now. Near the beginning of the answer @Andrew D. Hwang cites is: More precisely, a subset $D$ of $\mathbb R$ can be the discontinuity set for some derivative if and only if $D$ is an $F_{\sigma}$ first category (i.e. an $F_{\sigma}$ meager) subset of $\mathbb R.$ The rationals are such a set -- $F_{\sigma}$ because a countable union of singleton sets each of which is a closed set; first category because a countable union of singleton sets each of which is a nowhere dense set -- so the answer is YES. Maybe for an answer someone can give an explicit example. Jul 18 at 16:08
• Since $\Bbb{Q}$ is $G_{\delta}$ , we can produce a function which is discontinuous only on $\Bbb{Q}$. --- This is not correctly worded, and probably should be revised to one of the following two versions: "Since $\Bbb{Q}$ is $F_{\sigma}$ , we can produce a function which is discontinuous only on $\Bbb{Q}$." OR "Since $\Bbb{R}\setminus \Bbb{Q}$ is $G_{\delta}$ , we can produce a function which is continuous only on $\Bbb{R}\setminus \Bbb{Q}$." Jul 18 at 16:17
• Sorry $\Bbb{Q}$ is an $F_{\sigma}$ set. I have made a mistake there. I will fix immediately. Jul 18 at 16:20
• @Andreas: That does not work because $f$ has jump discontinuities at the rational numbers, so it can not be the derivative of some function $F$ (it does not have the intermediate value property). Jul 18 at 17:51

The following result is quoted in How discontinuous can a derivative be?, with references to several proofs:

A subset $$D$$ of $$\mathbb R$$ can be the discontinuity set for some derivative if and only if $$D$$ is an $$F_{\sigma}$$ first category (i.e. an $$F_{\sigma}$$ meager) subset of $$\mathbb R.$$

One of the references which is also online available is

$$\Bbb Q$$ surely is an $$F_{\sigma}$$ first category set, so there does exist a differentiable function $$f$$ on $$\Bbb R$$ such that $$f'$$ is discontinuous exactly on $$\Bbb Q$$. Inspecting the proof from the Bruckner/Leonard article, one can construct such a function for this set:

Let $$h: \Bbb R \to \Bbb R$$ be a function with the following properties:

• $$h$$ is differentiable everywhere.
• $$h'$$ is continuous everywhere except at $$x=0$$.
• $$h$$ and $$h'$$ are bounded.

(For example, $$h(x) = x^2 \sin(1/x)$$ for $$x\ne 0$$, $$h(0) = 0$$.)

Now let $$(q_n)$$ be an enumeration of the rational numbers, and define $$f: \Bbb R \to \Bbb R$$ as $$f(x) = \sum_{n=1}^\infty 3^{-n} h(x-q_n) \, .$$

Both $$\sum_{n=1}^\infty 3^{-n} h(x-q_n)$$ and $$\sum_{n=1}^\infty 3^{-n} h'(x-q_n)$$ converge uniformly on $$\Bbb R$$, so that $$f$$ is differentiable with $$f'(x) = \sum_{n=1}^\infty 3^{-n} h'(x-q_n) \, .$$

$$f'$$ is continuous at every irrational point $$x_0$$ because all functions $$3^{-n} h'(x-q_n)$$ are continuous at $$x_0$$ and the series for $$f'$$ converges uniformly.

And $$f'$$ is discontinuous at every $$q_m \in \Bbb Q$$ because in $$f'(x) = 3^{-m} h'(x-q_m) + \sum_{n \ne m} 3^{-n} h'(x-q_n) \, .$$ the first term on the right is discontinuous at $$q_m$$, whereas the second term is continuous at $$q_m$$, again as a uniformly convergent series of functions which are all continuous at $$q_m$$.

• Very interesting.... Jul 18 at 16:50
• @Lost in Space: For those interested, here is a freely available copy of the Bruckner/Leonard paper. Jul 18 at 19:50