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It is well known that an integrable function $u \colon \mathbb R^d \to \mathbb R$ is said to be of bounded variation iff the distributional gradient $Du$ is (representable by) a finite Radon measure, still denoted by $Du$.

Then it is also well known that the measure $Du$ can be decomposed into three terms, $D^{a}u, D^{j}u, D^c u$ resp. absolutely continuous part, jump part and Cantor part. Quite a plethora of fine results are present in the literature (see, e.g. the monography by Ambrosio-Fusco-Pallara also for the notation below, which is however quite standard). For instance, $D^au= \nabla u \mathscr L^d$ (being $\nabla u$ the approximate differential) and $D^j u = (u^+-u^-) \otimes \nu_{J_u} \mathscr H^{d-1}$.

On $D^c u$ little is present: it is always generically said that, as a measure, is something intermediate between the jump part $\mathscr H^{d-1}$ and the a.c. part $\mathscr L^d$. As an example, it is always considered the Cantor-Vitali staircase (the devil's function), whose derivative has only Cantor part and is exactly $\mathscr H^{\alpha}$, for $\alpha = \log_3 2$ which is - incidentally - the Hausdorff dimension of the standard Cantor set.

So here is my question:

Is it always true that $D^c u$ is (absolutely continuous w.r.t.) $\mathscr H^\alpha$ for suitable $\alpha \in (d-1, d)$? In other words, is $D^c u$ always an Hausdorff measure (up to a density) restricted to some Cantor-like set of certain dimension $\alpha \in (d-1,d)$?

That should have something to do with densities and Besicovitch Theorem... but I do not know.

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    $\begingroup$ What happens if you construct a Cantor-like set in $[0,1]$ by repeatedly cutting a middle interval, but such that each of the $2^n$ parts after the $n$-th cut has a length $\ell_n$ which tends to $0$ much faster than $(1/3)^n$ but much slower than $(1/3+\varepsilon)^n$ for any fixed $\varepsilon>0$? The resulting set would have H-dimension $\log2/\log3$ but would still be infinitely small w.r.t. H-measure of that dimension, so I think the derivative of its "staircase" would be a counterexample to your question. $\endgroup$
    – Gro-Tsen
    Commented Dec 28, 2017 at 10:54
  • $\begingroup$ @Gro-Tsen Thanks a lot for your comment, it makes perfectly sense. I think you are right: the Cantor set you constructed has still H-dimension $\alpha= \log 2 / \log 3$, nevertheless its staircase has not $\mathscr H^{\alpha}$ as derivative. Btw, one more question: the derivative of its staircase will still be some $\mathscr H^{\beta}$ (though not with $\beta = \alpha$), am I right? So in the end, is it true that the Cantor part of the derivative of a BV function is always (a.c. w.r.t. some) fractional Hausdorff measure? Thanks again. $\endgroup$
    – Romeo
    Commented Dec 28, 2017 at 13:30
  • $\begingroup$ No, I think the derivative of the staircase in the example I propose will not be $\mathscr{H}^\beta$ for any $\beta$: it is, in a certain sense, intermediate between $\mathscr{H}^\alpha$ for $\alpha=\log2/\log3$ and $\mathscr{H}^{\alpha-\varepsilon}$ for every $\varepsilon>0$, because the constructed Cantor set has measure $0$ for the H-measure of dimension $\alpha$ but infinity for the H-measure of any smaller dimension. The H-dimension is a very coarse classification, which can be indefinitely subdivided (if I understand correctly what is going on!). $\endgroup$
    – Gro-Tsen
    Commented Dec 28, 2017 at 13:59
  • $\begingroup$ @Gro-Tsen Uh, I see. That's really interesting how nasty Hausdorff measures (and dimension) can be! If you want to write your comments as an answer I will mark it as the accepted answer. Thanks a lot! $\endgroup$
    – Romeo
    Commented Dec 29, 2017 at 9:07
  • $\begingroup$ The thing is, I don't know how to properly justify all my claims about the Hausdorff measures of the Cantor-like set I described, which is why I wrote a comment (suggesting you consider it) rather than an answer. I'm fairly sure it works (and more generally, Cantor-like sets with appropriate growth rates can be used as counterexamples for a lot of things), but this is really not my specialty. $\endgroup$
    – Gro-Tsen
    Commented Dec 29, 2017 at 13:34

1 Answer 1

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Certainly not with a single $\alpha$, but it is tempting to decompose $D^cu$ into an integral $\int_{d-1}^d\mu_\alpha\ d\nu(\alpha)$ with $\mu_\alpha$ having a density with respect to $\mathcal H^\alpha$ on an $\alpha$-dimensional set. I don't know if it is always possible.

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  • $\begingroup$ Thanks for your interest. Any ideas on references? And btw you say "certainly not with a single $\alpha$": can I ask you if you have a counterexample to disprove my claim? $\endgroup$
    – Romeo
    Commented Dec 25, 2017 at 16:53
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    $\begingroup$ A sum of two "devil's staircases" whose derivatives are carried on Cantor-like sets of different dimensions... $\endgroup$
    – Jean Duchon
    Commented Dec 25, 2017 at 16:57
  • $\begingroup$ Oh you are right, that is true. So it is interesting to see whether this decomposition into $\alpha$-dimensional sets can be performed... the only thing that comes to my mind is coarea formula, but that does not seem to help... Thanks again. $\endgroup$
    – Romeo
    Commented Dec 25, 2017 at 17:06
  • $\begingroup$ Or maybe one could work with the sets $E_\alpha = \{x: \lim_{r \to 0} r^{-\alpha}|Du|(B^r(x)) > 0\}$... does this make sense? Thanks. $\endgroup$
    – Romeo
    Commented Dec 25, 2017 at 17:13

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