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?
[Math] Cantor function, strictly increasing
cantor setreal-analysis
Related Solutions
Aside. Your item 3 sets an unattainable goal. If a continuous function $f$ satisfies $f'=0$ for all but countably many points of $\mathbb R$, then $f$ is a constant function. See Set of zeroes of the derivative of a pathological function.
Here is a more explicit construction than in the post link in comments. Fix $r\in (0,1/2)$. Let $f_0(x)=x$. Inductively define $f_{n+1}$ so that
- $f_{n+1} (x) = f_n(x)$ when $x\in 2^{-n}\mathbb Z$
- If $x\in 2^{-n}\mathbb Z$, let $f_{n+1}(x+2^{-n-1}) = r f_n(x) + (1-r) f_n(x+2^{-n})$.
- Now that $f_{n+1}$ has been defined on $2^{-n-1}\mathbb Z$, extend it to $\mathbb R$ by linear interpolation.
- Let $f=\lim f_n$.
Here is a piece of this function with $r=1/4$:
and for clarity, here is the family $f_0$,$f_1$,... shown together. All the construction does is replace each line segment of previous step with two; sort of like von Koch snowflake, but less pointy. (It's a monotonic version of the blancmange curve).
To check that the properties hold, you can proceed as follows:
- Since $\sup|f_{n+1}-f_{n}| \le 2^{-n}$, it follows that $f_n$ is a uniformly convergent sequence. So the limit is continuous.
- The values of $f$ at every dyadic grid $2^{-n} \mathbb Z$ are strictly increasing, by direct inspection. Since $f$ eventually stabilizes at dyadic rationals, it is strictly increasing.
- Being increasing, $f$ is differentiable almost everywhere. Let $x$ be a point of differentiability (note that $x$ is not a dyadic rational). Then $$f'(x) = \lim_{n\to\infty} 2^{n} (f_n(x_n+2^{-n})-f_n(x_n)) \tag{1}$$ where $x_n\in 2^{-n}\mathbb Z$ is such that $x\in (x_n,x_n+2^{-n})$. The expression inside the limit in (1) is a product of the form $r^{i}(1-r)^j$ where $i+j=n$. More precisely, $i$ and $j$ are the numbers of $1$s and $0$s in the first $n$ digits of the binary expansion of the fractional part of $x$. It follows that $f'(x)=0$ unless $x$ has a lot more $0$s than $1$s in the binary expansion; but the number of such points $x$ has zero measure.
(It may be more enjoyable to prove part 3 using the Law of Large Numbers from probability.)
Monotone in words means it can only increase or it can only decrease. And in that case for instance a strictly increasing sequence is monotone since it cannot decrease.
Best Answer
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.