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

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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

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    $\begingroup$ So for general $A,B$ we can't always conclude $A=B$ ? 😞 PS: I hope your French is doing well πŸ™‚. $\endgroup$
    – Maximilian Janisch
    Commented Apr 1, 2021 at 16:00
  • $\begingroup$ 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 :) $\endgroup$
    – sharpend
    Commented Apr 1, 2021 at 16:04
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    $\begingroup$ @DanieleTampieri Ohhh right, since $A-B$ could be any weak solution to the Laplace equation. Took me a while to see πŸ˜†! Thanks to both of you! $\endgroup$
    – Maximilian Janisch
    Commented Apr 1, 2021 at 16:18
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    $\begingroup$ 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. $\endgroup$
    – Alexandre Eremenko
    Commented Apr 1, 2021 at 17:18
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    $\begingroup$ @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 πŸ™‚ $\endgroup$
    – Maximilian Janisch
    Commented Apr 1, 2021 at 18:11

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