1
$\begingroup$

Let $f:\Omega \subset \mathbb{R}^N \to \mathbb{R}^N$ be a vector field such that $f \in BV(\Omega)$.

Suppose that $\mathrm{div} f$ is absolutely continuous with respect to the Lebesgue measure and that its density is in $L^\infty$.

What does this imply about the derivative of $f$? For example, about its Cantor part?


Update 1: Is it true that $\mathrm{div}\, f$ is equal to the trace of $D_S f$ (that is to the trace of the rank-one matrix $M$ such that $D_S f = M|D_S f|$)? Why?

Update 2: What is the form of the rank-one matrix $M$ such that $D_S f = M|D_S f|$? That is, what are its entries in general? What are its entries in the case $\mathrm{div} \, f$ is absolutely continuous with respect to the Lebesgue measure?

$\endgroup$

1 Answer 1

3
$\begingroup$

According to G. Alberti (Proc. Roy. Soc. Edinburgh Sect. A 123 (1993), no. 2, 239–274), the singular part of $Df$ is a rank-one measure $D_Sf$. This is true for every BV vector field. When ${\rm div}\,f$ is a.c. with respect to the Lebesgue measure, then the trace of $D_Sf$ vanishes. However, I don't see what kind of information is conveyed by the boundedness of the divergence.

Edit. By the way, ${\rm div}\,f$ is not equal to ${\rm Tr}\,D_Sf$ in general. The correct statement is that the singular part of the measure ${\rm div}\,f$ equals ${\rm Tr}\,D_Sf$.

With your notation, $M$ is rank-one $|D_Sf|$-almost everywhere. It can be written $ab^T$ for $|D_Sf|$-measurable vector fields $a,b$. When ${\rm div}\,f$ is Lebesgue-a.c., then $a\cdot b=0$, $|D_Sf|$-almost everywhere.

$\endgroup$
4
  • $\begingroup$ I see. That's very helpful. Where can I find a proof of those statements? For clarity, how do you write $ab^T$ as a matrix? And how do you write it if $a \cdot b = 0$? $\endgroup$
    – Riku
    Apr 11, 2019 at 13:02
  • $\begingroup$ @Riku. $ab^T$ is naturally the ran-one matrix with entries $a_ib_j$, and $a\cdot b$ is the standard notation for scalar product. $\endgroup$ Apr 11, 2019 at 14:11
  • $\begingroup$ Thank you for the clarification. Is it possible to obtain what $a_ib_j$ are in terms of $f$? Could you recommend a reference where these statements on the trace of $D_Sf$ are proved? $\endgroup$
    – Riku
    Apr 11, 2019 at 14:13
  • $\begingroup$ Also, I've asked a related question here mathoverflow.net/questions/327789/… that you might be interested in. $\endgroup$
    – Riku
    Apr 11, 2019 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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