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I'm interested to know if a piecewise-continuous monotone function from $\mathbb{R}$ to $\mathbb{R}$ is also piecewise-continuously differentiable. I mean piecewise in the sense that there is a finite or countably infinite and isolated set $S\subset\mathbb{R}$, and intervals covering $\mathbb{R}\setminus S$ such that on any of these intervals, continuity and continuous-differentiability are verified.

I know that

  • a monotone function is almost everywhere differentiable (this topic)
  • there are monotone and everywhere differentiable functions that are not continuously differentiable (this other topic)
  • almost everywhere differentiable may not imply almost everywhere continuously differentiable, the derivative can actually be nowhere continuous (this one)

It seems to me that this third reference is not exactly a counter-example as from what I understand, it features a sort of devil-staircase function and I guess this one is not piecewise-continuous? (possible accumulating sequence of discontinuities)

M. Boyet
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    The classical Devil's staircase, $c$, is continuous and differentiable on precisely the complement of Cantor dust, which is a union of intervals. Where $c$ is differentiable, $c'=0$, and so $c\in C^1(\text{there})$. But the Cantor set is not isolated (it's perfect). Did you mean to rule this out somehow? – Jacob Manaker Dec 24 '21 at 21:01
  • Indeed, I'm only looking at functions that are continuous on the complement of a finite/finitely countable and isolated set. It seems off to me to say that the Devil's staircase is piecewise-continuous, or even piecewise-constant. I'm not sure that there is a total consensus on what "piecewise-xxx" means? – M. Boyet Dec 24 '21 at 21:32
  • If you consider (the unique continuous extension of) $f(x) = x^2\sin(1/x)$ to be piecewise continuous (since it's in fact differentiable) but not to be piecewise continuously differentiable (which is reasonable, since the one-sided limits of $f'$ don't exist at $0$), then the examples from the first linked question are counterexamples...? – Andrew D. Hwang Dec 24 '21 at 22:49
  • This puzzles me, the derivative of $f$ is continuous on both $(-\infty,0)$ and $(0,\infty)$ so it should be piecewise-continuous. Doesn't requiring existence of one-sided limits give a stronger notion of "piecewise-continuity"? – M. Boyet Dec 25 '21 at 10:16
  • Everywhere I've seen piecewise-$C^1$ defined it includes existence of one-sided limits of the derivative; effectively we want to treat $f$ as being of class $C^1$ on finitely many compact intervals $I_k$ with disjoint interiors. (That is, for each $k$ there is an open neighborhood of $I_k$ to which $f$ extends as a $C^1$ function.) If that's not part of the definition here, that detail may be worthwhile to add to the question body. – Andrew D. Hwang Dec 25 '21 at 16:01

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