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This question suddenly came to my mind (May be very silly question...)

Suppose $f$ is a continuous function over $\mathbb{R}$ and twice differentiable. Can $f$ ever has uncountable number of Local maxima or minima? (Twice differentiablity can be ignored for finitely many points).

I was thinking of counter-example (obviously), but could not find one.(I am really really bad at finding counter-example actually!)

I thought about the example $x^2\sin\Big(\dfrac{1}{x}\Big)$ when $x\neq 0$, and $0$, when $x=0$, but this function is not twice differentiable everywhere.

Added:

Note: Twice differentiablity can be ignored for finitely many points (Because I think if I impose this condition the problem will be interesting...)

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marked as duplicate by Chris Culter, Ben, Sahiba Arora, Community Aug 23 '17 at 13:19

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  • $\begingroup$ And has countably many extrema. $\endgroup$ – Yves Daoust Aug 23 '17 at 8:00
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    $\begingroup$ Your question nearly amounts to asking if a once differentiable function can have uncountably many roots. (But constant zero sections mut be ruled out.) $\endgroup$ – Yves Daoust Aug 23 '17 at 8:01
  • $\begingroup$ well yes... that's right!!! @YvesDaoust $\endgroup$ – MAN-MADE Aug 23 '17 at 8:02
  • $\begingroup$ @Arthur: that function is defined by an algorithmic procedure, from a function with countable extrema. I'd be surprised it leads to uncountabiity. $\endgroup$ – Yves Daoust Aug 23 '17 at 8:10
  • $\begingroup$ Note that this is not possible in $\mathbb C$, since a complex valued function with uncountably many zeroes must be identically zero by a version of the identity theorem. But that doesn't mean it isn't possible in $\mathbb R$. $\endgroup$ – wythagoras Aug 23 '17 at 8:13