Let $u\in BV(\Omega)$ be a function of bounded variation where $\Omega\subset \mathbb R^N$ is open bounded with smooth boundary. We use $Du$ to denote the weak derivative of $u$. (So $Du$ is a Radon measure), and we use $\nabla u$ to denote the absolutely continuous part of $Du$. We also use $S_u$ to denote the jump set of $u$ and we assume $\mathcal H^{N-1}(S_u)<\infty$.

Now let $u_\delta$ be a sequence of $BV$ function such that $u_\delta\to u$ weakly in $BV$ and $S_{u_\delta}\to S_u$ in the sense of Hausdorff distance.

My question: would it possible to modify $u$ to obtain $\tilde u_\delta$ such that $S_{\tilde u_\delta}=S_{u_\delta}$ and $\tilde u_\delta\to u$ in $L^1$ and $$ \int_{\Omega}|\nabla \tilde u_\delta|dx\to \int_\Omega |\nabla u|dx $$ as $\delta\to 0$.

Thank you!

  • $\begingroup$ Assuming that you want to “modify” the sequence $u_\delta$, why wouldn't $u_\delta =u$ for all $\delta$ do? Note that you make no assumptions about the quality of the approximating sequence. I somehow think that the Cantor function a.k.a. Devil's staircase on $\Omega=(0,1)$ would be a counterexample to the question you actually wanted to ask. $\endgroup$ – Manfred Sauter Mar 20 '18 at 19:42

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