# Does there exists a differentiable function $f$ on $\mathbb R$ whose derived function $f'$ is discontinuous on $\mathbb Q$ and continuous elsewhere?

Recently I found a problem which asks:

Does there exists a differentiable function $$f$$ on $$\mathbb R$$ whose derived function $$f'$$ is discontinuous on $$\mathbb Q$$ and continuous elsewhere?

More generally given any $$F_\sigma$$ set ,does there exist a differentiable function on $$\mathbb R$$ whose derivative has discontinuity only on that set and continuous elsewhere?

I attempted to make a function whose derivative $$f'(x)=t(x)$$ where $$t(x)$$ is the extended thomae function(thomae function extended for $$\mathbb R$$ instead of $$[0,1]$$). But my question is does the function $$t(x)$$ have an antiderivative on $$\mathbb R$$? I have not yet studied Riemann integrability,so I cannot conclude anything about it.

Yes:

Start with the standard $$h(t)=\begin{cases}t^2\sin(1/t),&(t\ne0), \\0,&(t=0).\end{cases}$$ So $$h$$ is differentiable and $$h'$$ is continuous except at $$0$$. Since $$h'$$ is locally bounded there exists a differentiable function $$g$$ with $$g(t)=h(t)$$ for $$|t|\le1$$ and such that $$g$$ and $$g'$$ are bounded.

Say $$\Bbb Q=\{r_1,r_2,\dots\}$$. Let $$f(t)=\sum 2^{-n}g(t-r_n).$$It follows that $$f$$ is differentiable and $$f'(t)=\sum 2^{-n}g'(t-r_n),$$since the last sum is uniformly convergent (cf. baby Rudin Thm 7.17.). It's clear that $$f'$$ is continuous at $$t$$ if $$t$$ is irrational, again by uniform convergence.

And $$f'$$ is discontinuous at $$t$$ if $$t$$ is rational. Details for that: Say $$t=r_n$$. Write $$f=f_1+f_2,$$where $$f_1(t)=2^{-n}g(t-r_n).$$Then as above, uniform convergence shows that $$f_2'$$ is continuous at $$r_n$$; since $$f_1'$$ is discontinuous there so is $$f$$.

## Note

No, the Thomae function $$f$$ does not have an antiderivative. But there's a major gap in the explanation for this in various comments: It's clear that if $$g(y)-g(x)=\int_x^yf$$ then $$g$$ is constant, hence $$g'\ne f$$. But it's not clear why $$g'=f$$ would imply $$g(x)-g(y)=\int_y^x f$$, since after all $$f$$ is not continuous. Possibly one could justify this using some fancy version of FTC.

Edit. In fact it's easy to show that if $$g$$ is differentiable and $$g'$$ is Riemann integrable then $$g(x)-g(y)=\int_y^x g'$$; I was forgetting this. So the argument in those comments is fine, although probably someone might have mentioned the bit about Riemann integrability.

Anyway, there's a simple argument without FTC:

The point being that although a derivative need not be continuous, it can't be "too discontinuous". For example it's well known that a derivative cannot have a jump discontinuity. That's not quite enough here, but:

Lemma. If $$g:\Bbb R\to\Bbb R$$ is differentiable then $$\limsup_{t\to0}g'(t)\ge g'(0)$$.

Proof: It's an easy exercise from the definitions to show there exists a sequence $$t_n$$ decreasing to zero such that $$\frac{g(t_n)-g(t_{n+1})}{t_n-t_{n+1}}\to g'(0).$$So MVT shows here exists a sequence $$s_n\to0$$ (with $$s_n>0$$) such that $$g'(s_n)\to g'(0).$$

Otoh if $$f$$ is the Thomae function then $$\limsup_{t\to0}f(t)So the lemma shows that $$f$$ is not a derivative.

• C. Utlrich Does thomae function have an antiderivative over $\mathbb R$. Jan 5 '20 at 2:32
• @KishalaySarkar: Thomae function can't be the derivative of any function. If it were so then we would have $$g(x) - g(y) =\int_{y} ^{x} f(t) \, dt$$ where $f$ is Thomae function. Since the integral is $0$ , $g$ is constant and $g'$ is identically $0$ and thus does not equal $f$. Jan 5 '20 at 6:12
• @ParamanandSingh Since $f$ is not continuous, why would $g'=f$ imply $g(x)-g(y)=\int_y^x f$? Jan 5 '20 at 11:06
• @KishalaySarkar No. See edit for a correct proof of this. Jan 5 '20 at 11:25
• There is a version of FTC which does not require continuity. See math.stackexchange.com/a/2149700/72031 Jan 5 '20 at 12:32