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?




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**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?



BV function with absolutely continuous divergence
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.


