Let's consider in dimension $d\geq 3$ the Newton/riesz potential $f=I_2[g]$ $$ f(x)=\int_{R^d}\frac{1}{|x-y|^{d-2}}g(y)dy, $$ which solves $-\Delta f=g$ (up to positive normalizing constants, which I shall ignore).

The classical Hardy-Littlewood-Sobolev inequality guarantees that $$ |f|_{L^q(R^d)}\leq C|g|_{L^p(R^d)},\qquad q=\frac{dp}{d-2p} $$ as soon as $g\in L^p(R^d)$ for some $p\in (1,d/2)$.

If $g$ is only in $L^1$ the naive guess $|f|_{L^{d/(d-2)}}\leq C|g|_{L^1}$ fails, but we know that the singular integral converges for almost every $x\in R^d$ and that there is a weak type estimate $$ \forall \lambda\geq 0, \qquad\mathcal{L}(\{x:\,|f|(x)\geq \lambda\})\leq C\left(\frac{|g|_{L^1}}{\lambda}\right)^{d/(d-2)} $$ ($\mathcal{L}$ denoting the usual Lebesgue measure on $R^d$).

**My question is the following**: what can we say if $g$ is in the slightly better space $g\in L^1\log L^1$, i-e if $|g|.|\log(|g|)|\in L^1(R^d)$? I don't think we can expect $f\in L^{d/(d-2)+\varepsilon}$ for any $\varepsilon>0$, but maybe $f\in L^{d/(d-2)}\log L^1$ or something like that? I would expect at least some improvement and in particular $f\in L^{d/(d-2)}$, but maybe not.

Thank you in advance for your comments and input.

**PS:** I wrote my question in dimension $d\geq 3$ for simplicity, but I'm also interested in the borderline case $d=2$ when the Poisson kernel $1/|x-y|^{d-2}$ is replaced by $-\log|x-y|$.