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Let $D$ be a constant coefficients linear differential operator on complex valued functions on $\mathbb{R}^n$. Let $\tilde D=\mathbb{F}^{-1}\circ D\circ \mathbb{F}$ be its conjugation with the Fourier transform, thus $\tilde D$ is a polynomial on $\mathbb{R}^n$.

Let us assume that $\tilde D$ does not vanish. Does it tell something on non-zero solutions of the equation $Du=0$?

A basic example is $D=-\Delta +1$.

Remark. As far as I understand it follows that $u$ does not belong to the Schwatz space of tempered growth distributions.

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    $\begingroup$ You are right: if such a $u$ belongs to ${\cal S}'(R^n)$, then $u=0$. $\endgroup$
    – Giorgio Metafune
    Commented Dec 10, 2020 at 16:03
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    $\begingroup$ On the other hand, without growth restrictions at infinity, no conclusions can be drawn (for example $D=-d^2/dx^2+1$ has the solutions $u=e^{\pm x}$). $\endgroup$
    – Christian Remling
    Commented Dec 10, 2020 at 16:47
  • $\begingroup$ @ChristianRemling : What is wrong with the exponent? $\endgroup$
    – asv
    Commented Dec 10, 2020 at 19:58

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