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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 !

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2 Answers 2

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.

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The last part of the argument doesn't seem correct. The function $f(x) = 1/2$ for $x\in[0,1/2)$ and $f(x) = 2x-1/2$ seems to be a counterexample to the last claim. –  Glen Wheeler May 30 '11 at 20:19
    
I modified the answer a bit : I used the fact that $f$ takes every value on $[0,1] $, but I guess my answer might still be incomplete, since I need to justify why the complement of the Cantor set gets mapped onto a set of measure zero, which is , I guess, intuitively clear. –  Mathmath May 31 '11 at 13:35
    
The complement of the Cantor set is a countable union of intervals, and $f$ is constant on each of these intervals. So $f(C^c)$ is a countable set. –  Jason Swanson Jun 29 '11 at 21:50

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 $0=f(0)<f(1)=1$.

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