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Call a function $f: \mathbb R+ \to \mathbb R$ essentially $C^\infty$ if there exists a sequence $f_n$ $(n \geq 0)$ such that each $f_n$ is differentiable a.e., $f_0 = f$ a.e., and $f_n’$ is equal to $f_{n+1}$ a.e.

Given a essentially $C^\infty$ function $f$, and two associated sequences $f_n$ and $g_n$, it is true that necessarily $f_n = g_n$ a.e - so to every essentially $C^\infty$ function $f$, we can assign a unique sequence of equivalence classes $[f_n]$.

Call a essentially $C^\infty$ function h essentially anti-Cauchy if for every $M > 0$ there exists some $N > 0$ such that $|h’| > M$ a.e. on $(N, \infty]$.

For every subset $S$ of the naturals, does there exist a essentially $C^\infty$ function $f$ such that $f_n$ is essentially anti-Cauchy exactly when $n$ is in $S$?

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  • $\begingroup$ why anti-Cauchy rather than anti-Lipschitz? $\endgroup$
    – YCor
    Commented May 15, 2019 at 7:26
  • $\begingroup$ Note: one difficulty in the definition is that "to be a.e. differentiable" is not invariant under a.e. equality. This is a little issue (already when defining a.e. twice differentiable), which maybe is the reason the definition is formulated this way rather than directly introducing iterated derivatives. Indeed, for $f$ a.e. differentiable, to say that $f$ is a.e. twice differentiable does not a priori mean that there exists $g$ a.e. equal to $f$ such that $g'$ is a.e. differentiable. $\endgroup$
    – YCor
    Commented May 15, 2019 at 7:34
  • $\begingroup$ Yep, I think you nailed the difficulty, and hence the convoluted definition $\endgroup$
    – James Baxter
    Commented May 15, 2019 at 7:44

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