The notations are mainly those of Functions of Bounded Variation and Free Discontinuity Problems by Ambrosio, Fusco and Pallara.
Hi,
In a problem I am considering, I have reached the following conclusion ($\Omega$ being an open bounded subset of $\mathbb{R}^2$ with smooth boundary): \begin{align*} z=\nabla \Phi \hspace{0.2cm}{\mbox{in $L^1(\Omega,\mathbb{R}^2)$}}, \hspace{0.2cm}{\mbox{so $z=\nabla \Phi$ a.e.}}, \end{align*} with the additional assumptions that $z\in SBV(\Omega,\mathbb{R}^2)$ (class of functions with special bounded variation) and $\Phi \in W^{1,\infty}(\Omega,\mathbb{R})$. By definition of a vector-valued bounded variation function, we conclude that $\nabla \Phi \in BV(\Omega,\mathbb{R}^2)$ (yielding subsequently that $\Phi \in BH(\Omega)$ (class of functions with bounded Hessian)) with $Dz=D(\nabla \Phi)$ ($2\times 2$ matrices of measures).
According to the Lebesgue decomposition theorem, $Dz=\nabla z dx+D^jz$, $D^jz$ denoting the jump part of $Dz$ (single component of the singular part $D^sz$ of $Dz$ since $z\in SBV(\Omega,\mathbb{R}^2)$). Similarly, $D^2\Phi=\nabla^2\Phi dx+D^s(\nabla \Phi)=\nabla^2\Phi dx+D^j(\nabla \Phi)+D^c(\nabla \Phi)$ (Cantor part for the latter component).
My question is probably obvious but can I straightforwardly conclude that $\nabla^2 \Phi=\nabla z$ a.e., $D^jz=D^j(\nabla \Phi)$ and $D^c(\nabla \Phi)=0$, resulting in $\Phi\in SBH(\Omega)$.
(I believe this is true by uniqueness of the Lebesgue decomposition but I would like to be sure).
Many thanks.