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Determine the existence of function $f$ such that $f$ is defined for all $x$ in $\mathbb{R}$, $f$ is differentiable, $f'(x) = 0$ for all rational number $x$, and $f'(x) \ne 0$ for all irrational number $x$.

I found examples where $f'(x) = 0$ for all rational number $x$ but $f$ is not constant : Link 1 Link 2

But these do not provide whether there exists an irrational number $\alpha$ such that $f'(\alpha) = 0$. Is this possible?

EDIT : It is impossible since the zero-set of a Pompeiu derivative must be a residual set. That is, the elements in the zero-set should be uncountable. However, since $\mathbb{Q}$, which is the zero-set in this case, is countable, it is a contradiction.

EDIT 2 : What if we change the conditions to $f'(x) = 0$ for all irrational number $x$ and $f'(x) \ne 0$ for all rational number $x$?

The answer is still a No. Since $f'(x) \ne 0$ for all rational numbers, we can find an arbitrary rational number $q$ such that $f'(q) \ne 0$. Pick an arbitrary irrational number $p$ that is smaller than $q$; then by Darboux's Theorem the derivative of $f$ should take all values from $0$ to $f'(q)$. That is, for all $t \in (0, f'(q))$ or $t \in (f'(q), 0)$ there exists $c \in (p, q)$ such that $f'(c) = t$.

Since $f'(x) = 0$ for all irrational number $x$, $c$ is rational.

Meanwhile, there are countable rational numbers in the interval $(p, q)$, but there are uncountable values(real values) between $0$ and $f'(q)$. This is a contradiction since we just discovered a surjection sending a countable set to an uncountable set - which is impossible.

So, both cases don't work.

Vue
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  • I admit I have difficulties to understand the function in the link that is non-constant but with zero-derivate on a dense set. Maybe, you can summarize why and how this works because it is a very counterintuitive result. – Peter Jan 25 '23 at 18:35
  • No, there does not exist a differentiable function $f\colon \mathbb{R}\rightarrow \mathbb{R}$ such that $f'(x)=0$ for $x\in \mathbb{Q}$ and $f'(x)\neq0$ for $x\notin \mathbb{Q}$. The derivative of any such function would be an example of a so-called Pompeiu derivative. The zero set of a Pompeiu derivative is a dense set which is the countable intersection of open sets. In particular, the zero set is residual, hence uncountable. The rationals are countable, however. – Max Demirdilek Jan 25 '23 at 21:27
  • If such a function, say $f$, existed, one could by a suitable tranlsation, however, find a function $g$ with these properties such that $g(\alpha)=0$ for some irrational $\alpha$. Namely, pick any $\alpha \in \mathbb{R}\setminus \mathbb{Q}$. Then $g=f-f(\alpha)$ has the desired properties. – Max Demirdilek Jan 25 '23 at 21:36
  • I'll post a more detailed answer if I find the time. – Max Demirdilek Jan 25 '23 at 21:37
  • @M.C. What do you mean by $f - f(\alpha)$? I think $g$ should satisfy $g'(\alpha) = 0$, not $g(\alpha ) = 0$. (I made a typo in my question.) I am looking forward to your elaboration. – Vue Jan 26 '23 at 01:51
  • @M.C. I do understand that such function $f$ cannot exist, by the property of a Pompeiu derivative. – Vue Jan 26 '23 at 02:02
  • Some results that are more precise than given in @M.C.'s first comment are in my answer to Set of zeroes of the derivative of a pathological function, which incidentally is within your "Link 1" (Link 1 $\rightarrow$ unique MSE link in comments $\rightarrow$ unique MSE link in comments). – Dave L. Renfro Jan 26 '23 at 08:21
  • @Vue With your edit the answer to you question is certainly negative. I mean, you ask for a function $f$ such that $f'(\alpha)=0$ and $f'(\alpha)\neq 0$ for some irrational number $\alpha$. – Max Demirdilek Jan 26 '23 at 08:37
  • Possibly of interest are my answers to Smallest positive zero of Weierstrass nowhere differentiable function AND Level sets of a Weierstrass nowhere-differentiable function. I had forgotten about these two answers when I posted my earlier comment. – Dave L. Renfro Jan 26 '23 at 10:11
  • @M.C. I think you misunderstood my question. I was asking for function $f$ where $f'(x) = 0$ only for rational numbers. The answers in the link show examples of functions that their derivative becomes zero on rational numbers but not constant; but no word on whether $f'(\alpha)$ can be zero for irrational number $\alpha$. – Vue Jan 26 '23 at 14:18
  • I think @M.C.'s answer works. I actually found something similar, so I'll add that in my question. – Vue Jan 26 '23 at 14:21

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