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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?

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    $f'$ maps a set of irrational numbers to an interval.2012-06-22

2 Answers 2

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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")

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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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    I have removed an inappropriate comment.2012-07-04