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For any $k\in\mathbb{N}$, the space $BV^k(\mathbb{R}^n)$ is defined as the set of all functions $f\in L^1(\mathbb{R}^n)$ such that there exists a Radon masure $D^\alpha f$ and for any $\phi\in C^\infty_c(\mathbb{R}^d)$, $$\int_{\mathbb{R}^d} f D^\alpha \phi = (-1)^{|\alpha|} \int_{\mathbb{R}^n}\phi d D^\alpha f$$ where $\alpha$ is a multi-index satisfying $|\alpha|\le k$. In particular when $k=1$, $BV^1$ is the space of all functions with bounded variation.

Suppose that $\Omega$ is a bounded open subset of $\mathbb{R}^n$, $n\ge 2$. What smoothness conditions on the boundary of $\Omega$ would ensure that $\chi_\Omega\in BV^2(\mathbb{R}^n)$?

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    $\begingroup$ By approximating $\chi$ by mollified versions and using lower semicontinuity of the variation, I think it can be shown that no characteristic function of a bounded open set can be of bounded second order variation. $\endgroup$ Commented Nov 9 at 6:39
  • $\begingroup$ $\chi_\Omega$ never belongs to $BV^2$, even when $\Omega$ is a round ball. $\endgroup$ Commented Nov 9 at 11:49
  • $\begingroup$ A "correct" question is when $\chi_\Omega\in BV^1$, and the answer is probably in terms of $(n-1)$-Hausdorff measure on $\partial\Omega$. $\endgroup$ Commented Nov 9 at 12:00
  • $\begingroup$ @NateRiver Thanks for your comment. The lower semicontinuity only proves that the second order variation is less than or equal to infinity. Could you please elaborate your argument how one can prove your claim? $\endgroup$ Commented Nov 9 at 16:02
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    $\begingroup$ @Ribhu: If $\Omega$ has a nice boundary, you can consider test functions of small support near a piece of the boundary, and we are effectively dealing with the case where the boundary consists of pieces of hyperplanes. Then a typical contribution to a first order derivative looks like $\int_{\mathbb R^{n-1}} \varphi (0,y)\, dy$. If you apply this distribution to $\partial\varphi/\partial x_1$, then it will obviously not be possible to control its value in terms of $\|\varphi\|_{\infty}$ only, so this is not a measure. $\endgroup$ Commented Nov 9 at 18:52

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