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I'm trying to construct a fuction described as follows: $f:[0,1]\rightarrow R$ such that $f'(x)=0$ almost everywhere,f has to be continuous and strictly increasing. (I'd also conlude that this functions is not absolutely continuous)
The part in bracket is easy to prove.
I'm in troubles with constructing the function:
I thought about considering the function as an infinite sum of sucession of increasing and conitnuous function, I considered the Cantor-Vitali functions which is defined on $[0,1]$ and is continuous and incresing (not strictly).
So $f(x)=\sum_{k=0}^{\infty} 2^{-k}\phi(3^{-k}x)$ where $2^{-k}$ is there to makes the sum converging and $\phi$ is the Cantor-Vitali function.
The sum convergethe function in continuous (as sum of) and is defined as asked.But I'm in trouble while proving that is strictly increasing.Honestly it seems to be but I don't know how to prove it.
I know that $0\leq x \leq y\leq 1$ always exist a k such that $\phi(3^{-k}x)\leq \phi(3^{-k}y)$ but then I stucked.
I'm looking for help in proving this part.

Laura
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1 Answers1

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By $\phi$ we denote Cantor-Vitali function. Let $\{(a_n,b_n):n\in\mathbb{N}\}$ be the set of all intervals in $[0,1]$ with rational endpoints. Define $$ f_n(x)=2^{-n}\phi\left(\frac{x-a_n}{b_n-a_n}\right)\qquad\qquad f(x)=\sum\limits_{n=1}^{\infty}f_n(x) $$ I think you can show that it is continuous and have zero derivative almost everywhere. As for strict monotonicity consider $0\leq x_1<x_2\leq 1$ and find interval $(a_n,b_n)$ such that $(a_n,b_n)\subset(x_1,x_2)$, then $$ f(x_2)-f(x_1)\geq f(b_n)-f(a_n)\geq f_n(b_n)-f_n(a_n)=2^{-n}>0 $$ So $f$ is strictly monotone.

Norbert
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  • Did you mean $n^{-2}$ or $2^{-n}$ I think could work in both ways..or not? – Laura Dec 04 '12 at 11:38
  • @Laura Both ways, I'll edit my answer to follow your choice. – Norbert Dec 04 '12 at 11:48
  • Thank you, this approach is more similar to the ones used by my teacher so I think he'd appreciate. – Laura Dec 04 '12 at 12:17
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    Just to be sure: were you thinking of a pointwise differentiation like that (http://people.math.sc.edu/howard/Notes/fubini.pdf) to prove that the derivative of f is zero a.e. ? – charmd Apr 07 '19 at 12:54
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    @CharlesMadeline, yes – Norbert Apr 07 '19 at 17:33
  • Can someone explain the last equality? I believe $f_n(b_n)=2^{-n}$, but if $0<x_1<1$ is irrational, then for every given choice of $n$ we can find $k$ such that $x_1< a_k < a_n$, whence $f(a_n)\geq f_k(a_n)>0$. – QuartelQuartz Mar 05 '24 at 15:44
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    @QuartelQuartz there was a typo. Fixed – Norbert Mar 05 '24 at 15:55