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Is there a function $f\colon \Bbb R\to \Bbb R$ in $C^\infty$ such that the set $$D=\{y\in \operatorname{Im}(f):\exists x\in f^{-1}(y)\text{ with }f'(x)=0\}$$ is uncountable?

This question arose when thinking about the regular values of a smooth map into a differentiable manifold. Sard's Theorem says that $D$ has measure zero, so a candidate here will probably be exotic.

It is not too hard to construct such a function where $D$ is dense: take a listing of the rationals and a function which has a plateau between $n$ and $n+1$ at the height of the $n$-th rational.

I will preemptively warn that this question is not about finding a nontrivial function whose derivative vanishes on an uncountable set. Such a function can be constructed by requiring $f(x)=0$ on a cantor set and extending smoothly.

pancini
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    Related. In the comment it is shown that $f(D)$ cannot be the Cantor set. The answer actually implies that $f(D)$ should have Hausdorff dimension zero. An example of Hausdorff dimension zero uncountable set is given here. –  Feb 09 '18 at 02:47
  • I meant $D$ in the previous comment. –  Feb 09 '18 at 02:53
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    If $f(x)$ is the example you preemptively warned us about, why won't the function $g(x)=\int_0^xf(t)dt$ do what you want? (I suppose $f(x)\gt0$ when $x$ is not in the Cantor set.) – bof Feb 09 '18 at 03:09
  • @bof I have just flown over the link. It doesn't look like that the result ensures $f(x) > 0$ for $x$ outside the cantor set. – user251257 Feb 09 '18 at 03:54
  • @user251257: See https://mathoverflow.net/questions/179445/non-zero-smooth-functions-vanishing-on-a-cantor-set in which I gave an explicit construction of a smooth function vanishing precisely on the Cantor set. – Nate Eldredge Feb 09 '18 at 04:31
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    @bof: Indeed, that seems like it would work nicely: since $f > 0$ almost everywhere, the function $g$ is strictly increasing, so $D$ is an injective image of the Cantor set and hence uncountable. Care to post as an answer? – Nate Eldredge Feb 09 '18 at 04:35
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    @NateEldredge OK, done. Thanks for the reference. – bof Feb 09 '18 at 05:31

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Let $f(x)$ be a $C^\infty$ function which vanishes on the Cantor set and is positive everywhere else, e.g., the function constructed by Nate Eldredge in this Math Overflow answer. Then the function $$g(x)=\int_0^xf(t)dt$$ is a strictly increasing $C^\infty$ function whose derivative vanishes precisely on the Cantor set, whence its set $D$ of critical values is a bijective image of the Cantor set and so has cardinality $\mathfrak c.$

bof
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  • Thanks; I guess I'm still a confused how this doesn't contradict Jason Ma's comment. It might be the case that the image of the cantor set is an uncountable "thin cantor set" of Hausdorff dimension zero, but now I'm out of my depth. – pancini Feb 09 '18 at 07:01
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    I guess you must be right, $D$ must be a Cantor set of Hausdorff dimension zero. The fact that it's a continuous image of the standard Cantor set doesn't tell us much, all Cantor sets are homeomorphic. But don't take my word for anything, I don't know much about Hausdorff dimension. – bof Feb 09 '18 at 07:17