Let $f \in C^\infty_c(\mathbb R^n)$, we know $$(Af)(x) := (f * \Gamma)(x) - (f * \Gamma)(0)$$ solves the Poisson's equation on $\mathbb R^n$ $$\Delta u = f.$$ where $\Gamma(x)$ is the fundamental solution of the Laplcace equation.
This means the operator $A$ defined above is a right inverse to the Laplacian operator. Of course there are many other right inverses, but I think the operator $A$ is somewhat a "canonical" right inverse.
For general $f\in C^\infty(\mathbb R^n)$, the expression $f * \Gamma$ does not make sense. However, I believe (at least intuitively) there are still many right inverses to Laplacian operator in this case. And I guess $$(Bf)(x) := \lim_{R\to \infty} \Big((f\chi_{B_R} * \Gamma)(x) - (f\chi_{B_R} * \Gamma)(0)\Big) $$ is well-defined and $B$ is again a "canonical" right inverse to Laplacian. Can I actually prove it?
This may be a silly question. So I'm very sorry if it wasted your time.