Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ a differentiable function such that $f'(x)=0$ for all $x\in\mathbb{Q}.$ $f$ is a constant function?

  • $\begingroup$ Are you asking if $f$ must be a constant function, or if a constant function satisfies the criterion? $\endgroup$ – jbowman Jun 22 '12 at 17:33
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    $\begingroup$ @jbowman: Undoubtedly the former. $\endgroup$ – Brian M. Scott Jun 22 '12 at 17:35
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    $\begingroup$ @BrianM.Scott: One hopes so, but I've been surprised a time or two! $\endgroup$ – jbowman Jun 22 '12 at 17:38
  • $\begingroup$ If $f$ is differentiable on $\mathbb{R}$, then $f'(x)$ exists on $\mathbb{R}$, and if $f'$ is discontinuous at $x=a$, then it must be an essential discontinuity. That is, there cannot be a jump discontinuity there. But I'm having trouble making that into a proof that $f$ has to be constant. $\endgroup$ – Shaun Ault Jun 22 '12 at 18:08
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    $\begingroup$ This question is related: math.stackexchange.com/questions/151931/… $\endgroup$ – j.p. Jun 22 '12 at 18:18

No, such a function is not necessarily constant.

At the bottom of page 351 of Everywhere Differentiable, Nowhere Monotone Functions, Katznelson and Stromberg give the following theorem:

Let $A$ and $B$ be disjoint countable subsets of $\mathbb{R}$. Then there exists an everywhere differentiable function $F: \mathbb{R} \to \mathbb{R}$ satisfying

  • $F'(a) = 1$ for all $a \in A$,
  • $F'(b) < 1$ for all $b \in B$,
  • $0 < F'(x) \leq 1$ for all $x \in \mathbb{R}$.

Choosing $A = \mathbb{Q}$ and an arbitrary (nonempty*) countable set $B \subset \mathbb{R} \setminus \mathbb{Q}$, we get an everywhere differentiable function $F$ with $F'(q) = 1$ when $q \in \mathbb{Q}$. So if we define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = F(x) - x$, this then is the desired function satisfying $f'(q) = F'(q) - 1 = 1 - 1 = 0$ for $q \in \mathbb{Q}$, and $f$ is not constant (or else its derivative would be zero everywhere by the mean value theorem).

*(in case you take "countable" to mean either "countably infinite" or "finite")


The problem with Vandermonde's proof (from Katznelson and Stromberg) is that it assumes that both sets A and B are countable. While the rationals are countable, the irrationals are not. The function is constant because of the surprising fact the derivative exists. Because the rationals are countable, the discontinuities of the function (if any) are countable; next, because the derivative exists at all these points we know that using the limit definition of the derivative that the function is continuous. Finally, because the derivative is zero at these points, we know (because the function is continuous) that the derivative is zero everywhere; therefore, the function must be constant (or at least a piecewise function with a countable number of piecewise definitions, each of which is constant but not necessarily the same value.... for example, on one open interval the function has one constant value and on another it has another....)

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    $\begingroup$ (1) I did not (and could not) take $B$ to contain all irrationals. Just one such point where $-1 < f'(b) < 0$ will assure $f(x)$ is non-constant. (2) We're given the value of $f'(x)$ only on $\mathbb{Q}$ but do know it exists everywhere, so $f$ has no discontinuities at all. (3) I don't know whose argument your criticism is about, but I just re-read and everything looks fine. Can you be more specific? $\endgroup$ – Vandermonde Jun 26 '12 at 19:04
  • $\begingroup$ I have removed an inappropriate comment. $\endgroup$ – Zev Chonoles Jul 4 '12 at 18:42

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