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Let $\Omega \subset \mathbb R^d$ with $d \geq 2$ (I am mostly interested in the case when $\Omega$ is the unit ball). A vector field in $L^p(\Omega,\mathbb R^d)$ is called a divergence measure field if its distributional divergence is a Radon measure. The space of divergence measure fields over $\Omega$ is denoted by $\mathcal D \mathcal M^p(\Omega, \mathbb R^d)$.

My question seems trivial, but I couldn't find an answer anywhere.

Question: Let $\mathcal C(\Omega,\mathbb R^d)$ be the space of continuous vector fields over $\Omega$. Is it true that $\mathcal C(\Omega,\mathbb R^d) \subset \mathcal D \mathcal M^p(\Omega)$?

Discussion: The result clearly fails in one dimension, since then $\mathcal D \mathcal M^p(\Omega) = BV(\Omega) \cap L^p(\Omega)$ (with a ${}_{loc}$ if $\Omega$ is unbounded) and there exist continuous functions that are not in $BV$. However, in more than two dimensions the inclusion $BV(\Omega,\mathbb R^d) \cap L^p(\Omega,\mathbb R^d) \subset \mathcal D \mathcal M^p(\Omega,\mathbb R^d)$ is proper. For example, $$ g(x,y) = \Big(\sin\big(\frac{1}{x-y}\big), \sin\big(\frac{1}{x-y}\big)\Big) \in \mathcal D\mathcal M^\infty(\mathbb R^2,\mathbb R^2) \setminus BV_{\mathrm{loc}}(\mathbb R^2,\mathbb R^2). $$

There are quite a few papers on continuous solutions of \begin{equation}\label{eq2} \text{div}\, v = F, \tag{1} \end{equation} where $F$ is a distibution. The closest I came to some sort of an answer is Theorem 3.7 (see also Definitions 2.1 & 2.3) in de Pauw and Pfeffer, Distributions for which div v = F has a continuous solution, 2007. It says that a continuous solution of \eqref{eq2} exists if and only if for any compactly supported sequence of test functions $\varphi_i \in \mathcal D(\Omega, \mathcal R^d)$ it holds \begin{equation}\label{eq3} \lim_i |\varphi_i|_1 = 0 \quad \text{and} \quad \sup_i |\nabla \varphi_i|_1 < \infty \quad \implies \quad \lim_i F(\varphi_i) = 0. \tag{2} \end{equation} A consequence would be that if there exists a distribution satisfying \eqref{eq3} which is not a Radon measure, then the inclusion I'm after would be false. However, I don't know if \eqref{eq3} implies that $F$ is a Radon measure (perhaps under additional assumptions on $\Omega$, such as compactness).

Any help will be appreciated.

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    $\begingroup$ Identifying $\mathbb R^d\simeq\mathbb R\times\mathbb R^{d-1}$, to get a counterexample for $d\ge 2$, why don't you just take $(x,y)\mapsto(f(x)\,\chi(y),0)$, where the smooth $\chi$ has compact support and $f$ is "your favourite" continuous function not having bounded variation? $\endgroup$
    – TaQ
    Commented Sep 2, 2020 at 3:42
  • $\begingroup$ @TaQ true. I guess you even don't need to have a $\chi$, you can extend any scalar $f(x)$ to a vector field $(f(x),0)$. I'm trying to think how to exclude this trivial case. $\endgroup$ Commented Sep 2, 2020 at 9:58
  • $\begingroup$ @TaQ to exclude your counterexample, one might restrict fields in $\mathcal C(\Omega,\mathbb R^n)$ to gradient fields, i.e. to consider the quotient space $\mathcal C(\Omega,\mathbb R^n)/\{g \colon \text{div }g \equiv 0\}$. This doesn't exclude the trivial case $(x,y) \mapsto (f(x),0)$, though $\endgroup$ Commented Sep 2, 2020 at 10:03

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