Suppose $y(x)$ is continuous and $y'(x)=0$ has uncountable many solutions but $y(x)$ is not constant on any interval. Is this possible?
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Wouldn't taking something like $y'(x) = d$ where $d$ is distance between $x$ and the closest point of Cantor set do the trick? (Just an idea, didn't make rigorous proof yet). – Abstraction May 25 '16 at 08:52
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What if require it to be $C^k$, or $C^\infty$? – May 25 '16 at 08:54
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I would usually recommend Counterexamples in Analysis for this kind of question but it isn't in there. My copy is getting updated... – Matthew Towers May 25 '16 at 12:07
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Related: Find a Continuous Function with Cantor Set Level Sets – Dave L. Renfro May 25 '16 at 13:59
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Take a Cantor set $C\subseteq [0,1]$ and let $d(x)$ be the distance function from $x$ to $C$. Then $(d(x))^2$ is differentiable with derivative vanishing at all points of $C$ hence uncountably many.
Mikhail Katz
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1By $d^2(x)$ do you mean $d(d(x))$ or $(d(x))^2$? My brain doesn't like $d(d(x))$. – Yakk May 25 '16 at 13:42
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Good point. I like differentials including second-order ones, but in this case they are irrelevant :-) @Yakk – Mikhail Katz May 25 '16 at 13:47
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$(d(x))^2$ isn't differentiable at the midpoints of the intervals making up the complement of $C$. You need to modify the function around these points. – Daniel Fischer May 31 '16 at 12:54
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The OP did not require the function to be differentiable everywhere. – Mikhail Katz May 31 '16 at 15:43
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Yes, it is true. A classic example of such an odd function is the Cantor function. It is defined like this: for $x$ in $[0,1]$
1.Express x in base 3.
2.If x contains a 1, replace every digit after the first 1 by 0.
3.Replace all 2s with 1s.
4.Interpret the result as a binary number. The result is c(x).
This function has derivative $0$ almost everywhere (i.e. the places where it doesn't have $0$ derivative has measure $0$) and yet somehow manages to increase in value from $0$ at $x=0$ to $1$ at $x=1$.
There is plenty written about the Cantor function. Perhaps wikipedia would be the best starting place: https://en.wikipedia.org/wiki/Cantor_function
Josh R
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3Cantor function is constant on $({1 \over 3},{2 \over 3})$, isn't it? – Abstraction May 25 '16 at 09:09
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1oh yes I didn't see the OP's last condition was a bit different from usual :( perhaps then this page could be useful: http://math.stackexchange.com/questions/250628/constructing-a-strictly-increasing-function-with-zero-derivatives – Josh R May 25 '16 at 09:27