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Let $f$ be a differentiable function on $\mathbb{R}$ such that

$$f'(x)=0,\forall x\in \mathbb Q.$$

Prove or disprove that

$$f(x)=c$$

for some constant $c$. I've heard this problem is true, but I'm not sure. Can you prove it or provide a counterexample?

Auxiliary question: I wonder the converse of it, if $f′=0$ for all irrational, can we say that $f$ is constant?

Seirios
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math110
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  • I've edited your question. Is this what you meant to be asking? – davidlowryduda May 20 '14 at 08:04
  • Hello,@mixedmath, is $f(x)=c(constant)$ – math110 May 20 '14 at 08:05
  • It should be. I am unable to prove it rigorously. – evil999man May 20 '14 at 08:08
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    I think it's true from the density of the rational numbers on the real line. since obviously $f$ is continuous on every rational point, you can get that it's continuous all over the real line. – Snufsan May 20 '14 at 08:33
  • When we assume that $f$ is not constant, $f'$ is discontinious for infinitly many points on a closed interval. (this may cause contradiction). I do not know "Is there any theorem related countablty of discontinious points ?" – mesel May 20 '14 at 08:54
  • You may want to consider the fundamental theorem of Lebesgue integral calculus, which requires $f'(x)=0$ almost everywhere to get to your result. However, $\mathbb{Q}$ is far away from almost everywhere.. – Easy May 20 '14 at 09:08
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    This looks almost like a counter example: http://en.m.wikipedia.org/wiki/Pompeiu_derivative#Pompeiu.27s_construction. $f$ is differentiable on $[0,1]$ and $f'(x)=0$ on a dense subset. – Martin R May 20 '14 at 09:17
  • My intuition is that this is false. The Cantor function isn't a counterexample, but it's a lot like one. – user2357112 May 20 '14 at 09:18
  • It's false : if someone can read some french the answer is here http://www.daniel-saada.eu/fichiers/09-Fonctions_strictement_croissantes.pdf –  May 20 '14 at 09:47
  • @Edwin: Are there given counter example ? (I can not read french) – mesel May 20 '14 at 10:03
  • @mesel Yes, I add it as an answer but the proof it non-trivial.. –  May 20 '14 at 10:05

2 Answers2

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[I use a construction given by Rooij and Schikhof in A second course on real functions, Example 13.2.]

Let $\Omega$ be the collection of all continuous functions $\omega : \mathbb{R} \to (0, + \infty)$ having the following property: for all $a<b \in \mathbb{R}$, $$ \left| \frac{1}{b-a} \int_a^b \omega(x) dx \right| \leq 4 \omega(a).$$

Let $\alpha_1,\beta_1,\alpha_2,\beta_2,\dots$ be pairwise distinct real numbers. Then the authors show that one can construct $\omega_1,\omega_2, \dots \in \Omega$ such that for each $n \in \mathbb{N}$ we have, writing $f_n:= \omega_1+ \cdots + \omega_n$:

$$f_n \leq \left( \frac{n}{n+1} \right)^2, \ \ f_n(\alpha_i) \geq \left( \frac{n-1}{n} \right)^2, \ \ f_n(\beta_i) < \frac{n}{n+1} \cdot \frac{i}{i+1}.$$

Afterward, they show that $\displaystyle F(x):= \sum\limits_{i \geq 1} \int_0^x \omega_i(t)dt$ is well-defined and differentiable with $F'(x)= \sum\limits_{i \geq 1} \omega_i(x)$. In particular, because $F'(x)= \lim\limits_{n \geq + \infty} f_n(x)$, we have $$F'(\alpha_i)=1, \ \ F'(\beta_i) \leq \frac{i}{i+1}.$$

Let $f$ denote the function given by the construction above with $\{\alpha_1,\alpha_2,\dots \}= \mathbb{Q}$ and $\{\beta_1= \sqrt{2}, \beta_2, \dots \}$. Let $g$ denote the function given by the consruction above with $\{A_1= \sqrt{2},A_2, \dots\}= \mathbb{Q} \cup \{\sqrt{2} \}$ and $\{B_1, B_2, \dots\}$.

Let $G:=f-g$. Then $G'(q)=0$ for all $q \in \mathbb{Q}$ and $G'(\sqrt{2}) \leq \frac{1}{2}-1=- \frac{1}{2} <0$. In particular, from the mean value theorem, we deduce that $G$ is not constant since its derivate does not vanish everywhere.

For your second question, it is a consequence of Darboux's theorem that a function is constant if its derivative vanishes on $\mathbb{R} \backslash \mathbb{Q}$.

Indeed, $f'(\mathbb{R})$ has to be an interval $I$, but $f'(\mathbb{R} \backslash \mathbb{Q})= \{0\}$, so $I$ is in fact countable: necessarily, $I$ is the singleton $\{0\}$, that is $f'$ vanish everywhere. Now, it may be deduced that $f$ is constant from the mean value theorem.

J. W. Tanner
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Seirios
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  • I wonder the converse of it, if $f'=0$ for all irrational, can we say that $f$ is constant? – mesel May 20 '14 at 10:08
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    I had an argument in my answer. I also edited your question in order to add your second question. – Seirios May 20 '14 at 12:03
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It's false, you can find an answer here. (French)

The proposed function is $$ g(x)= \inf\{d(x,E_k)^{1/k}: k\in \mathbb{N}\} $$ where $E_k$ is the closed discrete set $\{p/2^k : p\in \mathbb{Z}\}$

Seirios
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  • I wonder the converse of it, if $f'=0$ for all irrational, can we say that $f$ is constant? – mesel May 20 '14 at 10:09