Global estimate to an L1 function whose Laplacian is a bounded measure Pretty simple question:
Suppose that $u \in L^1(\mathbb{R}^N)$ is such that $\Delta u \in \mathcal{M}(\mathbb{R}^N)$ (i.e., $\Delta u$ is a bounded Radon measure). Does $\nabla u \in L^1(\mathbb{R}^N)$?
In a bounded domain with zero Dirichlet boundary condition the answer is YES (see [Ponce, A. - Topics in Elliptic PDEs and Measure Theory]). But to unbounded domain I don't know.
 A: More a comment than an answer, but too long for a comment. First a comment on Michael Renardy's remark: there is no homogeneous function in $L^1(\mathbb R^N)$ so the first assumption is not satisfied. I believe that the answer to your question is related to  the Gagliardo-Nirenberg Inequality which says
$$
\Vert w\Vert_{L^{\frac{N}{N-1}}(\mathbb R^N)}\le C_N\Vert \nabla w\Vert_{L^{1}(\mathbb R^N)}.
$$
In fact that inequality can be applied to a $BV$ function $w$ where the rhs stands for the total mass of the measure $\nabla w$. Then if $\Delta u$ is a measure with a finite total mass, then $\nabla u$ should belong to $L^{\frac{N}{N-1}}$: that point is not completely obvious, because singular integrals are not bounded on $L^1$. However, this is what the scaling of the problem is suggesting. Then following the scaling, you will get that $u$ belongs to $L^p$ with
$$
\frac{N-1}{N}-\frac 1p=\frac1N\quad \text{i.e.}\quad p=\frac N{N-2}.
$$
A: The answer is yes. Here is a sketch of the argument.

Claim 1: Suppose that $u \in L^1$ and $\Delta u \in \mathcal M$. Let $f = u - \Delta u$. Then $f \in \mathcal M$ and $u = \mathcal B_2 * v$, where $\mathcal{B}_\alpha$ is the Bessel potential kernel.

Formally, this is clear, as $\mathcal{B}_2$ is the inverse of $(\operatorname{Id} - \Delta)$. I did not attempt to write a rigorous proof, but this should not be very difficult, using, for example, the ideas from Stein's Singular integrals and differentiability properties of functions.

Claim 2: The gradient of $\mathcal B_2$ is integrable.

This follows easily from the explicit expression for $\mathcal B_2$.
It follows that $$\|\nabla u\|_1 = \|\nabla \mathcal (B_2 * f)\|_1 = \|(\nabla \mathcal B_2) * f\|_1 \leqslant \|\nabla \mathcal B_2\|_1 \|f\|_{\mathcal M} < \infty,$$
as desired.
