For $0<\alpha<d$, fractional integral operator $I_{\alpha}$ is defined by $$I_{\alpha}f(x)=\int_{\mathbb{R}^d} \frac{|f(y)|}{|x-y|^{d-\alpha}} dy$$ for any suitable function on $\mathbb{R}^d$. The important inequality about $I_{\alpha}$ is Hardy-Littlewood-Sobolev theorem, that is $I_{\alpha}$ is bounded from $L^p$ to $L^q$ if $\frac{1}{q}=\frac{1}{p}-\frac{\alpha}{d}$ where $1<p<\frac{d}{\alpha}$.
Let $d\geq 3$, $\alpha=2$, and $f$ be a function which have second continuously derivative and compactly supported on $\mathbb{R}^d$. Define $I_2(x)=\int_{\mathbb{R}^d} \frac{|f(y)|}{|x-y|^{d-2}} dy$. One can show that $$u(x)=\frac{1}{d(d-2)|B(0,1)|}I_2f(x)$$ satisfy the Poisson equation $$ -\Delta u=f.$$ By Hardy-Littlewood-Sobolev inequality, we get an estimate for $u$, $\|u\|_q\leq C \|f\|_p$ where $q=\frac{pd}{d-2p}$.
On the other hand, we also have weak type inequality for $I_{\alpha}$, that is there exist a constant $C>0$ such that $$\gamma^q |\{x\in \mathbb{R}^d: |I_{\alpha}f(x)|>\gamma\}|\leq C\|f\|_p^q.$$
Is there any relation between the weak type inequality for $I_{\alpha}$ and the solution of Poisson equation?