11
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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 Dec 28 '17 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 Dec 28 '17 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 Dec 28 '17 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 Dec 29 '17 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 Dec 29 '17 at 13:34
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Dec 25 '17 at 16:53
  • 2
    $\begingroup$ A sum of two "devil's staircases" whose derivatives are carried on Cantor-like sets of different dimensions... $\endgroup$ – Jean Duchon Dec 25 '17 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 Dec 25 '17 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 Dec 25 '17 at 17:13

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.