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Let $f:\mathbb Q\to\mathbb R$ be a uniformly continuous function and assume that $f'(x)=0$ for all $x\in\mathbb Q$. That $f$ is constant is obvious...and, as far as I can tell, unprovable. Please tell me that I'm wrong!

Casper
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    This is a great example of why the word "obvious" is dangerous in mathematics – Alex Mathers Jan 17 '17 at 07:34
  • Strongly related: https://math.stackexchange.com/questions/2039063/ – Watson Jan 17 '17 at 08:50
  • Moreover, we should notice that since $f : \Bbb Q \to \Bbb R$ is uniformly continuous, it extends to a unique continuous $g : \Bbb R \to \Bbb R$. However, I don't know why $g$ should be differentiable if $f$ is assumed to have a "rational derivative" at any point, i.e. the limit $\dfrac{f(x+h)-f(x)}{h}$ exists for all rational $x$, when the rational number $h$ tends to $0$. – Watson Jan 17 '17 at 09:06
  • There exist functions whose derivative vanishes on a dense $G_\delta$ set – Andres Mejia Feb 08 '17 at 08:06

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It is unprovable because there are counterexamples, such as Minkowski's question mark function (Wiki).

See also: Is there any function continuous in R and differentiable in rational numbers with zero derivative?

Upd: $?'(q) = 0 | \forall q \in \mathbb{Q}$ can be proven by taking $q_n \to q^+$ such that ${?(q_n) - ?(q) \over q_n - q} \to 0$ and $r_n \to q^-, {?(r_n) - ?(q) \over r_n - q} \to 0$ (since the function is monotonic, for any other sequence $x_n \to q^+$, ${?(x_n) - ?(q) \over x_n - q}$ can be "sandwiched" between two similar subsequences based on $q_n$; since it's continuous, pointwise limit value equals derivative value).

Let $q = [q_0; q_1, ..., q_k]$, then take $q_n = [q_0; q_1, ..., q_k, n]$ and see that $?(q_n) = ?(q) + (-1)^k2^{-(a_1+a_2+...+a_k+n-1)}$; $q_n = {an+b \over cn+d}$ where $q = {a \over c}$, $a,b,c,d$ don't depend on $n$. Then ${?(q_n) - ?(q) \over q_n - q} = (-1)^k{c(cn+d) \over (bc-ad)2^{a_1+...+a_k+n-1}} \sim n2^{-n} \to 0$. Taking $r_n = [q_0; q_1, ..., q_k -1, 1, n]$ concludes the proof.

Community
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Abstraction
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  • Wauw! That is super interesting! Can I bother you with a follow up question? Is there some way to strengthen the assumptions so that it does become true (without making the conditional trivial)? – Casper Jan 17 '17 at 08:24
  • Little question: why is Minkowski's question mark function uniformly continuous on $\Bbb R$? It is on every compact set, obviously, but on the whole real line I'm not sure. Thank you! – Watson Jan 17 '17 at 09:09
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    @Watson One can just take it on $[0,1]$ and then extend to $[0,2]$ by $f(x) = f(2-x)$ and then on whole $\mathbb{R}$ by $f(x) = f(x+2)$. Then for any given $\varepsilon>0$ $\delta$ you get for $[0,1]$ is enough everywhere else. – Abstraction Jan 17 '17 at 09:13
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    @Abstraction : you don't even have to extend it to the real line : Minkowski's question mark function is already defined on $\mathbb{R}$ and $1$-periodic. – charmd Jan 17 '17 at 09:28
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    @Casper One interesting statement is that if $f$ is analytic complex function, then it's constant on whole $\mathbb{C}$, not just real line. But of course this assumption is too strong. I'm not sure if it's enough to require $f'$ being defined everywhere - I have a feeling there still would be a counterexample, but can't build it (Minkowski function derivative isn't defined everywhere, for example see http://link.springer.com/article/10.1007/s10958-012-0750-2 ). – Abstraction Jan 17 '17 at 15:47
  • @Abstraction: Except for Wikipedia and other answers on this site, I have been unable to find a source that confirms what you say. I need to be able to cite such a source. Do you (or someone else) know of one? (I have asked this as a separate question here: http://math.stackexchange.com/questions/2126705/academic-reference-concerning-minkowskis-question-mark-function) – Casper Feb 03 '17 at 21:42
  • @Casper Statements should be obvious enough. Wiki even contains the reference for continuity statement; $?'(x) = 0 | x \in \mathbb{Q}, x = [a_0; a_1, ..., a_k]$ follows from that and $?(x_n) = ?(x) \pm 2^{-(s+n)}$ where $x_n = [a_0; a_1, ..., a_k, n]$ and thus $|{?(x_n) - ?(x) \over x_n - x}| \sim n2^{-n} \to 0$. I can edit my answer to elaborate on that if you think it's necessary. – Abstraction Feb 06 '17 at 10:04
  • @Abstraction: I would actually prefer to refer to proofs in the literature rather than put them in my own text, if possible. The citation on Wikipedia to Finch does not give me what I need. Finch only states that the derivative is zero almost everywhere, and I need sources for (1) that the derivative is zero on all rational numbers and (2) that on irrational numbers it is either 0 or undefined (or if you like $+\infty$). – Casper Feb 06 '17 at 11:01
  • @Abstraction: I have looked at some of the other papers listed on Wikipedia. They do not state these things explicitly. I suspect that the problem is that they are trivial corollaries of what they do state explicitly - trivial enough that they don't find them worth stating, but not trivial enough that I see it. – Casper Feb 06 '17 at 11:01
  • Update: I now have a reference for (2). But I still need one for (1). – Casper Feb 07 '17 at 05:25
  • @Casper elaborated on my comment in the answer. Can't help with a reference, sorry (even if I were to search for it, I would likely find a book in Russian which probably won't help you much). – Abstraction Feb 08 '17 at 08:03
  • @Abstraction: Thank you very much for your help! – Casper Feb 08 '17 at 17:49