For $\Omega\subset\mathbb{R}^N$ open and bounded, let $W^{1,p}(\Omega)$ denote the usual Sobolev space of $L^p(\Omega)$ functions with weak partial derivatives in $L^p(\Omega)$ and $W_0^{1,p}(\Omega)$ the closure of $\mathcal C^\infty_c(\Omega)$ in this space.

Let $E\subset \Omega$ be a Caccioppoli set (i.e. a set of finite perimeter in $\mathbb{R}^N$), $U\subset\Omega$ be open and $w\in W_0^{1,p}(\Omega)$. Suppose that $$ \int_E \operatorname{div}(\eta w) = 0 \qquad\forall\;\eta\in\mathcal C^1_c(U;\mathbb R^N).$$ Question: Does there exist a sequence $w_n\in\mathcal C^1_c(\Omega)$ with $\mathcal H^{N-1}(U\cap\partial^*E \cap [w_n\neq 0])=0$ that converges to $w$ in $W^{1,p}(\Omega)$?
(Here $[w_n\neq 0]=\{x\in\Omega:w_n(x)\ne 0\}$ and $\mathcal H^{N-1}$ is the $(N-1)$-dimensional Hausdorff-measure).

The assertion seems natural to me because for $w\in\mathcal C_c^1(\Omega)$ the equality implies (as $E$ is Caccioppoli) $$ \int_{\partial^*E} w\eta \nu_E = 0 \qquad\forall\;\eta\in\mathcal C^1_c(U;\mathbb R^N),$$ where $\nu_E$ is the inner normal of $E$ which exists $\mathcal{H}^{N-1}$-a.e. on the reduced boundary $\partial^*E$. This then implies that $w=0$ $\mathcal H^{N-1}$-a.e. on $\partial^*E$. The same holds by the divergence theorem if $E$ has Lipschitz boundary and $w$ is Sobolev because then the trace of $w$ on $\partial E$ is well defined, and hence there exists a sequence of $\mathcal C_c^1(\Omega)$ functions vanishing on $\partial E\cap U$ that approximate $w$ in $W^{1,p}(\Omega)$.

I couldn't find much on other cases, but maybe the following reference helps for the case when $w\in W_0^{1,p}(\Omega)\cap L^\infty(\Omega)$: Chen, Gui-Qiang; Torres, Monica, Divergence-measure fields, sets of finite perimeter, and conservation laws, Arch. Ration. Mech. Anal. 175, No. 2, 245-267 (2005). ZBL1073.35156.

Are the above arguments for $w\in\mathcal C^1_c(\Omega)$ or if $E$ has Lipschitz boundary correct? If so, what about the general case where $w$ is only in $W^{1,p}_0(\Omega)$ and $E$ is merely Caccioppoli?

  • $\begingroup$ Can you clarify what you mean by Caccioppoli set? $\endgroup$
    – timur
    Aug 30, 2012 at 14:43
  • $\begingroup$ A set $E\subseteq\mathbb R^n$ is a Caccioppoli set iff its characteristic function $\mathbb 1_E$ has bounded variation, iff there is a vector-valued radon measure $D\mathbb1_E$ with $$\int_{\mathbb R^n}\mathbb1_Ediv\varphi d\mathcal L^n=-\int_{\mathbb R^n}\varphi d(D\mathbb 1_E)$$ for all $\mathbb R^n$ valued compactly-supported $\mathcal C^1$-functions $\varphi$, where $\mathcal L^n$ is the n-dimensional Lebesgue measure. In this case $D\mathbb 1_E$ is the distributional derivation of $\mathbb 1_E$. $\endgroup$
    – Elgrimm
    Aug 31, 2012 at 19:03
  • $\begingroup$ Oh in the notation of the question $D\mathbb 1_E=\nu_E\cdot\mathcal H^{N-1}|_{\partial^*E}$, where $\mathcal H^{N-1}|_{\partial^*E}(A):=\mathcal H^{N-1}(\partial^*E\cap A)$ $\endgroup$
    – Elgrimm
    Sep 1, 2012 at 11:03

1 Answer 1


Firstly, the condition in your question is sensible as it implies that $w$ vanishes $\mathcal{H}^{N-1}$-a.e. on $U\cap\partial^*E$. Moreover, your arguments for the two cases where $w\in C^1_\mathrm{c}(\Omega)$ or $E$ has Lipschitz boundary are correct. I expect that also the case $p>N$ where $w$ has a continuous representative works out. However, in general it is too much to ask that the approximating sequence is both smooth and vanishing on $U\cap\partial^*E$ up to a $\mathcal{H}^{N-1}$-nullset.

No, you cannot expect such a sequence to exist if $p\le N$ and $w\in W^{1,p}_0(\Omega)\cap L^\infty(\Omega)$ and $E$ has finite perimeter in $\mathbb{R}^N$ (i.e. $E$ is Caccioppoli).

As the construction of the example is somewhat technical, I shall first sketch the argument. An open set of finite perimeter $E$ can have massive amounts (say positive Lebesgue measure) of topological boundary $\partial E$ while the essential or reduced boundary $\partial^*E$ is small and very regular. In fact, one can arrange that there exists a nontrivial Sobolev function $w\in W^{1,p}_0(\Omega)$ supported in $\partial E\setminus E$ that vanishes $\mathcal{H}^{N-1}$-a.e. on $\partial^*E$ as $\partial^*E$ is regular. Then $w$ satisfies the assumption in the question as it vanishes on $E$ (along with its gradient). Any continuous function that vanishes $\mathcal{H}^{N-1}$-a.e. on $\partial^*E$ actually vanishes everywhere on $\partial E$. So one cannot use such functions to approximate $w$ in $W^{1,p}(\Omega)$.

Example: It suffices to consider the case where $\Omega$ is the unit ball in $\mathbb{R}^N$ and $U=\Omega$. For simplicity let $p=2$ and $N\ge 2$, but the argument works without change for $p\in(1,N]$. We make use of the same construction as in my answer https://mathoverflow.net/a/295459. Let $E:=\bigcup_k B_k$ be a countable union of open balls with pairwise disjoint closures contained in $\Omega$, such that $E$ is dense in $\overline{\Omega}$ and $$\operatorname{cap}(E)\le\sum_k\operatorname{cap}(B_k)<\operatorname{cap}(\Omega).$$ Here for $A\subset\mathbb{R}^N$ we denote by $\operatorname{cap}(A)$ the Sobolev capacity $$ \operatorname{cap}(A)=\inf\{ \|u\|^p_{W^{1,p}} : u\in W^{1,p}(\mathbb{R}^N)\text{ and }u\ge1\text{ a.e. on neighbourhood of }A\}. $$

Then $\sum_k\mathcal{H}^{N-1}(\partial B_k)<\infty$ because of the above estimate on the capacity. (Alternatively, just choose the balls small enough.) It is readily checked that $E$ has finite perimeter (i.e. is Caccioppoli) and $\partial^*E = \partial^*E\cap U = \bigcup_k\partial B_k$, possibly up to a $\mathcal{H}^{N-1}$-nullset.

As in the linked answer (the function is called $u$ there), there exists a nontrivial bounded $w\in W^{1,p}(\mathbb{R}^N)$ supported in $K := \overline{\Omega}\setminus E$. As $\Omega$ has a regular boundary, it follows by standard trace theory that $w\in W^{1,p}_0(\Omega)$. Observe that $$\int_E\operatorname{div}(\eta w) = \int_\Omega \mathbf{1}_E(w\operatorname{div}(\eta) + \nabla w\cdot \eta)=0$$ for all $\eta\in C^1_\mathrm{c}(\Omega;\mathbb{R}^N)$ as both $w$ and $\nabla w$ vanish on $E$ (recall that $w$ is supported in $K$).

Suppose now that $w_n\in C^1_\mathrm{c}(\Omega)$ with $w_n=0$ $\mathcal{H}^{N-1}$-a.e. on $\partial^*E$. Then due to continuity and the structure of $\partial^*E$ one has $w_n=0$ everywhere on $\partial^*E\cup K$. As $w$ is nontrivial on $K$, the sequence $(w_n)$ cannot possibly converge to $w$ in $W^{1,p}(\Omega)$.

The above example shows that for general finite perimeter sets $E$ it is too much to ask that the approximating sequence is continuous and vanishes $\mathcal{H}^{N-1}$-a.e. on $\partial^*E$. I expect, however, that everything works out if one replaces the latter condition by something like $\mathcal{H}^{N-1}(U\cap\partial^*E\cap [w_n\ne 0])<\frac{1}{n}$ or $\|w_n\|_{L^1(U\cap\partial^*E,\mathcal{H}^{N-1})}<\frac{1}{n}$.

The assumptions in the question imply that $w$ (actually its approximately continuous representative) vanishes $\mathcal{H}^{N-1}$-a.e. on $U\cap\partial^* E$. I shall not go into details, but that claim follows from $$ \int_U \mathbf{1}_E w\operatorname{div}(\eta) = - \int_U \mathbf{1}_E\nabla w\cdot \eta, $$ taking the supremum over $\eta\in C^1_\mathrm{c}(U;\mathbb{R}^N)$ with $|\eta(x)|\le 1$ for all $x\in U$ on both sides, and using that $\mathbf{1}_E w\in\mathrm{SBV}(\mathbb{R}^{N})$ with the jump behaviour that one would expect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.