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.



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 $f$.


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 $$\limsup_{t\to0}f(t)<f(0).$$So the lemma shows that $f$ is not a derivative.

  • $\begingroup$ C. Utlrich Does thomae function have an antiderivative over $\mathbb R$. $\endgroup$ Jan 5 '20 at 2:32
  • 1
    $\begingroup$ @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 $f$. $\endgroup$
    – Paramanand Singh
    Jan 5 '20 at 6:12
  • $\begingroup$ @ParamanandSingh Since $f$ is not continuous, why would $g'=f$ imply $g(x)-g(y)=\int_y^x f$? $\endgroup$ Jan 5 '20 at 11:06
  • $\begingroup$ @KishalaySarkar No. See edit for a correct proof of this. $\endgroup$ Jan 5 '20 at 11:25
  • $\begingroup$ There is a version of FTC which does not require continuity. See math.stackexchange.com/a/2149700/72031 $\endgroup$
    – Paramanand Singh
    Jan 5 '20 at 12:32

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