Let $f(x)=(\nabla N *g)(x)$, where $N(x)=\frac{1}{|x|^{n-2}}$ for $n\geq 3$ is the Newtonian kernel, and $g\in L^1(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$. Then we can have that $f\in L^{\infty}$ by separating $\mathbb{R}^n$ into $|x-y|\leq r$ and $|x-y|>r,$ but is it possible to have $f\in L^1$ from these conditions?
My thoughts: I can only see that $f$ should have decay while I don't know how fast it can be. Or do we have an example that $f$ is not in $L^1$?
No, you should not expect $f$ to be in $L^1(\mathbb{R}^n)$. For example, let $g$ be $1$ on the unit ball and $0$ elsewhere. When $|x|$ is large, $f(x)$ is approximately $c\nabla (|x|^{2-n})$ since from a large distance, the ball works about the same as a point mass. So, $|f(x)|$ decays like $|x|^{1-n}$, which is too slow to be in $L^1$.