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On Wikipedia: https://en.wikipedia.org/wiki/Cantor_function it says that the cantor function is non-decreasing. But I am wodering, if we restrict our domain to the cantor-set, does then the function become strictly increasing?

user119615
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No. When you create the Cantor "devil staircase", call it $V(t)$, you fix the value of the function on all the open intervals that are removed in the construction of the Cantor set. Since $V(t)$ is continuous, the value at the end-points of each removed interval (which are included in the Cantor set) must be the same. Therefore, $V(t)$ is not strictly increasing.

Edit: $V(t)$ is not strictly increasing even if it is restricted to the cantor set, since the endpoints of the removed intervals belong to the Cantor set (although the set contains much more than those, of course), and $V(t)$ attains the same value at the two endpoints of each such interval.

However, it is possible to construct a strictly increasing function whose derivative is zero a.e.

Edit: if you're curious on how this function can be built, look here.

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bartgol
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  • Thank you, but did you see that I asked for when the domain is restricted to the cantor-set? As far as I know the cantor set does not contain any open intervals? Or did you take this into account? – user119615 Aug 13 '15 at 15:19
  • That is, my question is regarding the cantor function, when it is only defined on the cantor set, we don't look at what happens at the removed intervals, where the function is constant. – user119615 Aug 13 '15 at 15:21
  • Yes. And as a matter of fact, the end-point of the removed interval do belong to the Cantor set. – bartgol Aug 13 '15 at 15:21