Can $f$ ever has uncountable number of Local maxima or minima?

Asked Aug 23 '17 at 07:58

Active Aug 23 '17 at 08:27

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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...)

edited Aug 23 '17 at 08:27

asked Aug 23 '17 at 07:58

MAN-MADE

5,381

And has countably many extrema. – Aug 23 '17 at 08:00
3

Your question nearly amounts to asking if a once differentiable function can have uncountably many roots. (But constant zero sections mut be ruled out.) – Aug 23 '17 at 08:01
well yes... that's right!!! @YvesDaoust – MAN-MADE Aug 23 '17 at 08:02
@Arthur: that function is defined by an algorithmic procedure, from a function with countable extrema. I'd be surprised it leads to uncountabiity. – Aug 23 '17 at 08:10
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 $\mathbb R$. – wythagoras Aug 23 '17 at 08:13
@wythagoras how to prove that, any link may be.. – MAN-MADE Aug 23 '17 at 08:15
See: https://math.stackexchange.com/questions/1334557/entire-function-with-uncountably-many-zeros – wythagoras Aug 23 '17 at 08:17
why can't you pick a function so that it is constant in an interval and then fix the corner points to have differentiability? – Arash Aug 23 '17 at 08:28
Also Countability of local maxima on continuous real-valued functions – Arash Aug 23 '17 at 08:41
If a local extremum is strict, it is required to be unique in an open interval around it. That open set has positive width and so there can only be countably many in a unit interval, and thus on the real line. – Reinstate Monica Aug 23 '17 at 10:14

Can $f$ ever has uncountable number of Local maxima or minima?

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