Functions of bounded variation (BV) on $\mathbb{R}^d$ are functions with $$ \|u\|_{BV} := \|u\|_{L^1(\mathbb{R}^d)} + TV(u) $$ finite, where $$ TV(u) := \sup\left\{ \int_{\mathbb{R}^d}u(x)div(\phi)(x)dx \mid \phi\in C^1_c(\mathbb{R}^d,\mathbb{R}^d),\, \|\phi\|_{\infty}\leq 1 \right\}. $$ From this it follows that if $u\in BV$ then $\mu:=Du \in \mathcal{M}(\mathbb{R}^d,\mathbb{R}^d)$ with $\|\mu\|_{\mathcal{M}}=TV(u)$, i.e. $\mu$ is a vector of Radon measures with finite total variation equal to the variation of $u$. To the best of my knowledge, the converse is not true in general. I am mainly interested in cases that the measure $\mu$ is not absolutely continous with respect to the Lebesgue measure. I have looked in books like
- Variational Analysis in Sobolev and BV spaces, by Hedy Attouch et al.
- Measure theory and fine properties of functions, by Evans and Gariepy
to find if there are sufficient conditions, but I did not find any there. So my question is:
Question: What are sufficient conditions on $\mu\in\mathcal{M}(\mathbb{R}^d,\mathbb{R}^d)$ such that there is a function $u\in BV$ with $Du=\mu$?
Hunch: Maybe sufficient conditions can be found when $\mu$ is compactly supported. In that case, we can take a more general $\mu$ and apply a mollifier to get a sequence of measures $\mu$ such that $\mu_\epsilon$ is compactly supported. When we construct the corresponding $u_{\epsilon}$, it is easy to show that $TV(u_\epsilon)\leq \|\mu\|_{\mathcal{M}}$. What then remains is to show that $u_{\epsilon}\to u$ strongly in $L^1$. Perhaps for this strong convergence some additional conditions are needed.
Side note: Whilst I could not find sufficient conditions for $BV$, I found if-and-only-if statements for $BV^*$. These need the growth conditions $\mu(B(x,r))\leq Cr^{d-1}$ (see e.g. Integral Inequalities of Poincaré and Wirtinger Type for BV Functions). The current version (3 May version) of ChatGPT gets thrown of by these results in trying to answer this question.
Similar question but different conditions: Radon measure as the derivative of a BV function
In the spirit of @gerw's comment, a sufficient (and necessary) condition for there to be a $u \in BV_{loc}(\mathbb{R}^d)$ with $Du = \mu$ would be that the distributional Jacobian of $\mu$ is symmetric, that is, $$\int_{\mathbb{R}^d} \partial_i \varphi\,d\mu_j = \int_{\mathbb{R}^d} \partial_j \varphi \,d\mu_i \quad \text{for all $\varphi \in C_c^{\infty}(\mathbb{R}^d)$ and $i,j \in \{1,\ldots,d\}$.}$$ Under this condition we can consider a family of standard mollifiers $(\eta_\epsilon)_\epsilon$ and define the smooth functions $$v_\epsilon:\mathbb{R}^d \to \mathbb{R}^d, \quad v_\epsilon(x) = (\eta_\epsilon * \mu)(x)=\int_{\mathbb{R}^d} \eta_{\epsilon}(x-y)\,d\mu(y).$$ It is then not hard to check that the Jacobian of $v_\epsilon$ is symmetric, so that we find a sequence $(u_\epsilon)_\epsilon \subset C^{\infty}(\mathbb{R}^d)$ with $\nabla u_\epsilon = v_\epsilon$. Now, we fix a $R>0$ and assume that each $u_\epsilon$ has zero average on the ball $B_R(0)$. By the Poincaré inequality we find that $$\|u_\epsilon\|_{L^1(B_R(0))} \leq C \|v_\epsilon\|_{L^1(B_R(0);\mathbb{R}^d)} \leq C\|\mu\|_{\mathcal{M}(\mathbb{R}^d;\mathbb{R}^d)}.$$ This shows that $(u_\epsilon)_\epsilon$ is bounded in $BV(B_R(0))$ so that up to a non-relabelled subsequence $u_\epsilon \to u$ in $L^1(B_R(0))$ for some $u \in BV(B_R(0))$. Since $\nabla u_\epsilon \mathcal{L}^n=v_\epsilon \mathcal{L}^n$ converges weak* to $\mu$, we must have $Du = \mu$. Because we can do this for any $R>0$, and the limit is unique up to constants, we may conclude there is some $u \in BV_{loc}(\mathbb{R}^d)$ with $Du=\mu$.
Finally, if we want to have $u \in BV(\mathbb{R}^d)$, it would be enough to assume that $\mu$ has compact support when $d \geq 2$. Indeed, then the $u$ constructed above would be constant outside a large ball, and we can subtract this constant to get an integrable function. (In the case $d=1$ the complement of a ball is not connected so the constant may be different on either side. To resolve this, one could additionally assume $\int_{\mathbb{R}} \,d\mu=0$.)