By Alberti rank-one theorem, could it be possible to reduce the study of a function $u \in BV(\mathbb{R}^N, \mathbb{R}^N)$ to the study of a function $\tilde{u} \in BV(\mathbb{R}^2, \mathbb{R}^2)$? At least as far as the derivative is concerned?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ You should be more specific about what you mean by "reduction". $\endgroup$– BazinCommented Apr 12, 2019 at 9:49
-
$\begingroup$ @Bazin I don't know precisely, I'd like to know about anything that can be done. I was thinking maybe one direction given by the "singular one" given by Alberti rank one and the other one $\mathbb{R}^{N-1}$ the orthogonal. $\endgroup$– user123457Commented Apr 12, 2019 at 17:59
-
$\begingroup$ As written in my answer on Mathoverflow, taking a $π΅π$ vector field on $\mathbb R^n$ with absolutely continuous divergence amounts essentially to deal with $π=\sum_{1β€πβ€n}π_j(π₯β²,π₯_n)\frac{\partial}{\partial x_j}$ where $π₯β²\in \mathbb R^{n-1}$ and $\frac{\partial a_1}{\partial x_n}$ Radon measure whereas all other entries of $π·π$ are absolutely continuous. Although it is not sufficient, it means that a good idea is to start with $π_1(π₯_1,π₯_2)\frac{\partial}{\partial x_1}$ in $\mathbb R^2.$ $\endgroup$– BazinCommented Apr 15, 2019 at 13:00
Add a comment
|