Let $S(\mathbb R^n)$ denote the space of all [Schwartz functions](https://en.wikipedia.org/wiki/Schwartz_space) 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](https://en.wikipedia.org/wiki/Poisson%27s_equation), which in essence means that, in the weak sense,
$$\Delta A=\Delta B,$$
where $\Delta$ is the (weak) [Laplacian](https://en.wikipedia.org/wiki/Laplacian). Can we conclude that $A=B$ ?




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*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 $$\Delta \phi=\tilde\phi.$$



Uniqueness of distributional solutions to the Poisson equation
You can conclude that $A - B$ is a harmonic polynomial: If you take the Fourier transform, you see that $|\xi|^2 (\hat{A} - \hat{B}) = 0$, so $\hat{A} - \hat{B}$ is supported at the origin. (All of this makes sense at the level of tempered distributions.) Therefore, $A - B$ is a polynomial. (This required the classification of distributions supported at a single point -- see Hormander, Vol. 1.)


If you know beforehand that, say, $A - B \in L^2$, then you can conclude that $A = B$.


-Dallas


