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I have the following question: I have a function $f: \mathbb R \to \mathbb R$ which is differentiable everywhere. I also have a set $G\subset\mathbb R$ which is dense in $\mathbb R$ and a $G_\delta$-set. I know that $f'(x)=0 \forall x\in G$.

Can I conclude that $f$ is constant?

The answer is yes if $f'$ is continuous, but unfortunately I don't know that.

Many thanks!

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Since the zero set is always a $G_\delta$, the question is whether it being dense implies that the function is constant. This is false, and a counterexample is known as a Pompeiu derivative.

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  • $\begingroup$ This function is not however differentiable everywhere, contrary to the assumption in the question. I suspect the question was just misstated though since it doesn't seem to be interesting if you assume that. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Nov 29, 2016 at 16:30
  • $\begingroup$ @R.. The first sentence of the link says that it's a derivative of an everywhere differentiable function. $\endgroup$
    – Cameron Zwarich
    Commented Nov 29, 2016 at 16:32
  • $\begingroup$ The WP article is sloppy; it says the derivative at some points "$=+\infty$" and papers over what that means. Maybe there are senses in which it makes sense to call this everywhere-differentiable anyway, but I would call it ae-differentiable. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Nov 29, 2016 at 16:35
  • $\begingroup$ The function they actually take in the end is the inverse of that one, which has finite derivative everywhere. $\endgroup$
    – Cameron Zwarich
    Commented Nov 29, 2016 at 16:56
  • $\begingroup$ @R if you read more carefully, $g(x)$ is the inverse of a function $f$ with finite derivative at every point, and $f'(x)=0$ iff $g'(f(x))=+\infty$. Of course "derivative" here has the usual meaning of limit of the incremental ratio in the extended real line. Quite a nice and well-written article, indeed. $\endgroup$
    – Pietro Majer
    Commented Nov 29, 2016 at 22:22

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