How to prove that the Cantor ternary function is not weakly differentiable?

Question

I am using the standard cantor ternary function $f$ here, as cited in this Wikipedia page.

It is an example of continuous, monotone increasing, but not strictly monotone increasing function with zero derivative almost everywhere. But how should I prove that its weak/distributional derivatives do not exist? I guess I start of with assuming that there exist $ g \in L^1_\text{loc}(R)$ such that $\int_R {f\phi'} = - \int_R{g\phi}$ for all $\phi\in C_c^\infty (R)$. And then I have to probably choose appropriate mollifiers $\phi_\epsilon$ and let $\epsilon \to 0$. But I am kind of stuck here; could you give me a detailed proof?

Also, is the derivative of $f$ a measure in the distributional sense?

Thank you !

Answer 1

I think I got an answer, please give me your opinion .

If possible, assume that the Cantor ternary function $f$ is weakly differentiable on $[0,1]$. Then the continuous function $f$ is absolutely continuous on $[0,1]$ ( any theory of PDE book has the proof: Sobolev functions $W^{1,p}(\text{interval }I)$ is AC on that interval $I$ for $p<\infty$), and hence maps sets of measure zero to sets of measure zero. But for the Cantor ternary function $f$, it maps the Cantor set to a set of measure 1, (since on the complement of the Cantor set, $f$ is constant,and $f$ takes every value in between $0$ and $1$ ), which is a contradiction.

Answer 2

Show that if $I\subset[0,1]$ is one of the intervals on which $f$ is constant, then $g$ (the presumed weak derivative of $f$) must be equal to 0 at almost every point of $I$. Do this by considering the various $\phi\in C^\infty$ vanishing outside of $I$. Since the union of these intervals has measure $1$, the derivative $g$ must vanish almost everywhere on $[0,1]$, which cannot be, because $0=f(0)<f(1)=1$.