# Relationship between continuous vector fields and divergence measure fields in dimension $\ge 2$

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 $$$$\label{eq2} \text{div}\, v = F, \tag{1}$$$$ where $$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)$$ it holds $$$$\label{eq3} \lim_i |\varphi_i|_1 = 0 \quad \text{and} \quad \sup_i |\nabla \varphi_i|_1 < \infty \quad \implies \quad \lim_i F(\varphi_i) = 0. \tag{2}$$$$ A consequence would be that if there exists a distribution satisfying \eqref{eq3} which is not a Radon measure, then the inclusion I'm after would be false. However, I don't know if \eqref{eq3} implies that $$F$$ is a Radon measure (perhaps under additional assumptions on $$\Omega$$, such as compactness).

Any help will be appreciated.

• 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?
– TaQ
Commented Sep 2, 2020 at 3:42
• @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. Commented Sep 2, 2020 at 9:58
• @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 Commented Sep 2, 2020 at 10:03