# Uniqueness of distributional solutions to the Poisson equation

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

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 $$A = B$$.
• So for general $A,B$ we can't always conclude $A=B$ ? 😞 PS: I hope your French is doing well 🙂. – Maximilian Janisch Apr 1 at 16:00
• Right, it just isn't true at that level of generality. ($A$ and $B$ could be different constants.) But at least you know what their difference must be. And thank you :) – sharpend Apr 1 at 16:04
• @DanieleTampieri Ohhh right, since $A-B$ could be any weak solution to the Laplace equation. Took me a while to see 😆! Thanks to both of you! – Maximilian Janisch Apr 1 at 16:18
• There is no polynomial, except 0, in the space $S(R^n)$, if I correctly understand that $S(R^n)$ is the space of Schwartz test functions. – Alexandre Eremenko Apr 1 at 17:18
• @AlexandreEremenko Yes absolutely, but $A-B$ can be of the type $A-B: S(\mathbb R^n)\to\mathbb R, \phi\mapsto\int_{\mathbb R^n} p\phi\,\mathrm dx$ for a polynomial $p$. I am pretty sure that that is what Dallas meant 🙂 – Maximilian Janisch Apr 1 at 18:11