# Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ a differentiable function such that $f'(x)=0$ for all $x\in\mathbb{Q}$ [closed]

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?

• Are you asking if $f$ must be a constant function, or if a constant function satisfies the criterion? Jun 22, 2012 at 17:33
• @jbowman: Undoubtedly the former. Jun 22, 2012 at 17:35
• @BrianM.Scott: One hopes so, but I've been surprised a time or two! Jun 22, 2012 at 17:38
• 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. Jun 22, 2012 at 18:08
• This question is related: math.stackexchange.com/questions/151931/…
– j.p.
Jun 22, 2012 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")