Let $S(\mathbb R^n)$ denote the space of all Schwartz functions on $\mathbb R^n$ equipped with the topology induced by the usual Schwartz semi-norms. Let $S(\mathbb R^n)^*$ denote its dual.

**My question.** Suppose that $A, B\in S(\mathbb R^n)^*$ both satisfy the same Poisson equation, which in essence means that, in the weak sense,
$$\Delta A=\Delta B,$$
where $\Delta$ is the (weak) Laplacian. Can we conclude that $A=B$ ?

*My idea.* Just following the definitions, we see that $\Delta A=\Delta B$ means that $\langle A,\Delta \phi\rangle = \langle B, \Delta\phi\rangle$ for all test functions $\phi\in S(\mathbb R^n)$. Can we conclude that $\langle A,\tilde\phi\rangle = \langle B, \tilde\phi\rangle$ for all test functions $\tilde\phi\in S(\mathbb R^n)$ ? I am asking because there exist $\tilde\phi\in S(\mathbb R^n)$ for which there is no $\phi\in S(\mathbb R^n)$ such that $$\Delta \phi=\tilde\phi.$$