Consider these different notions of continuity, listed in order of decreasing strength:

  • Continuously differentiable
  • Lipschitz continuous
  • Absolutely continuous
  • Uniformly continuous
  • Continuous

I was wondering how far up the list we have to go before it is ensured that a function $f:\mathbb R\to\mathbb R$ must be such that $f'(x_0)$ for an irrational $x_0$ can always be recovered from information about only the values of $f'$ for rational numbers, i.e., that $f'(x_0)=\lim_{x\to x_0}f'|_{\mathbb Q}(x)$ whenever the right-hand side of that equation is defined.

I know that uniform continuity is not sufficient: Minkowski's_question_mark_function is a counter-example (for all rational $x$, $f'(x)$ is $0$, so the RHS of the equation is also always $0$, but there are irrational $a$ such that the LHS is not). On the other hand, $f$ being continuously differentiable is obviously sufficient. But what about absolute and Lipschitz continuity?

  • $\begingroup$ Absolute continuous functions are characterized as those function having weak-derivative. In particular a continuous function $f$ is absolutely continuous whenever it has a "weak derivative" $g$, i.e. a function such that $f(x)= f(0)+ \int_0^x g(t) \ \mathrm{d}t$. $\endgroup$ – Crostul Jan 23 '17 at 18:36
  • $\begingroup$ @Crostul: But is the weak derivative guaranteed to have the property I'm after? $\endgroup$ – Casper Jan 23 '17 at 20:23

Continuously differentiable is enough, obviously.

Lipschitz continuous is not enough. Enumerate rational points as $q_1,q_2,\dots.$ Let $U=\bigcup_{n=1}^\infty(q_n-2^{-n}, q_n+2^{-n})$. This is an open set of measure at most $1$ which contains all rational points. Let $$f(x)=\begin{cases} m([0,x]\setminus U)\,\quad & x>0\\ 0,\quad & x\le 0\end{cases}$$ where $m$ is the Lebesgue measure. This is a nonconstant $1$-Lipschitz function. Its derivative is equal to $0$ at all rational points, by construction.

By the way, $f'(x)=1$ for a.e. $x\in (0,\infty)\setminus U$, by the Lebesgue density theorem.


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