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In this paper, it is written that Alberti’s rank says that the singular part $D^s u$ with respect to $\mathcal L^d$ of the distributional derivative $Du$ of a function $u \in BV_{loc}(\mathbb R^d; \mathbb R^m )$ can be written, in polar decomposition, as $D^s u = \xi \otimes \eta|D^s u|$.

Then "by a standard blow-up argument this implies that near to $|D^s u|$-a.e. point $x$ asymptotically $u(y)$ behaves like a function having a single non-zero component, parallel to $\eta(x) \in \mathbb{S}^{m-1}$, and depending on a single scalar variable, the component of $y$ along $\xi(x) \in \mathbb{S}^{d-1}$".

  1. What does the statement in quotes mean heuristically?
  2. How can it be proved rigorously (that is, could you outline the details of the "standard blow-up argument" mentioned above?
  3. Where can I find a picture to represent this situation?

A more generic question was asked in Meaning of Alberti rank-one theorem. A related issue is in Alberti rank-one theorem and reduction of the study of BV function to the two-dimensional case.

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1 Answer 1

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The answer to your questions 1. & 2. can be found in Theorem 3.95 of the book Ambrosio, Fusco, Pallara Functions of Bounded Variation and Free Discontinuity Problems.

Another very recent and excellent reference is the book by Rindler Calculus of Variations, paragraph 10.4 (where you also find an excellent picture).

It is not easy to sketch the proof: the underlying idea is to introduce a suitable rescaling of $u$ (see eq. (3.92) in the book by Ambrosio, Fusco and Pallara) for which you get automatically compactness. The study of the structure of the limit (which is the core of the Theorem) is rather delicate and relies on theorems about tangent measures and, of course, on Alberti's Rank One Theorem. In this way you are able to conclude that the blow up is a vector valued function which is "directed" in only one direction (say $\eta$) and depends only on the other direction (say $\xi$). This explanation is very poor and the best I can do is to refer you to the aforementioned books.

Concerning the picture to represent the situation, I suppose the best is if you try to work out the details of what is going on by yourself. A good toy-model to work with is a "piecewise defined" vector field $u=u(x,y)$ in $\mathbb R^2$ (e.g. something like $u_1(x,y)$ if $x<0$ and $u_2(x,y)$ otherwise: compute the jacobian matrix and find by yourself $\eta,\xi$). Of course this example shows the rank-one property for the jump part and not for the Cantor part (which is actually the crucial point of Alberti's Theorem but is also considerably more difficult to "visualize").

Hope this helps.

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  • $\begingroup$ Thank you. Could you sketch the proof of that theorem? And clarify what it means heuristically? Yes, visualizing the Cantor part is the difficulty. And I'd really appreciate some insight on that issue. $\endgroup$
    – user123457
    Commented Apr 24, 2019 at 20:08
  • $\begingroup$ On Mathoverflow there are other questions related to this point: mathoverflow.net/questions/328790/…, mathoverflow.net/questions/328493/…, mathoverflow.net/questions/328792/… $\endgroup$
    – user123457
    Commented Apr 24, 2019 at 20:09
  • $\begingroup$ @Rene I have added some informal comments and another reference. Hope this helps. $\endgroup$
    – Romeo
    Commented Apr 24, 2019 at 20:18
  • $\begingroup$ I've seen the picture, which is the same as the one on page 14 of iperpv2017.files.wordpress.com/2016/11/dephilippis.pdf. This leads me to echo another question (mathoverflow.net/questions/328779/…): what is a concrete example of $BV(\mathbb R^2)$ function with such Cantor type singular derivative? $\endgroup$
    – user123457
    Commented Apr 25, 2019 at 0:34

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