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I have (hopefully) a rather basic question about smooth elliptic partial differential equations.

Let $L$ be a linear elliptic differential operator with polynomial coefficients in $\mathbb{R}^n, n>1.$ Let $u\in L^{\infty}(\mathbb{R}^n)$ be such that it has compact support and $L(u)$ is supported on a finite set (as a distribution). Then is $u$ necessarily 0? I (naively) hope that the answer is yes, and perhaps could be proved by using some known asymptotics of the fundamental solution of $L$ near the support of $u.$ Any suggestions or references will be greatly appreciated.

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  • $\begingroup$ Could you clarify? If $u$ has compact support, then isn’t $Lu$ always a compactly supported distribution? $\endgroup$
    – Deane Yang
    Commented Jun 11, 2020 at 21:42
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    $\begingroup$ Right, but question is if $Lu$ can have a support that is a finite set, which is much stronger that having a compact support. $\endgroup$
    – Zamanyan
    Commented Jun 11, 2020 at 21:58
  • $\begingroup$ you probably want $n > 1$, else the tent function $\psi$ has $\psi''$ supported at exactly three points. // A minor nitpick about your phrasing: since your conclusion is $u \equiv 0$ a fortieri you cannot have $L(u)$ with non-empty support. Maybe better to say that $\mathrm{supp} L(u)$ is contained in a finite set. $\endgroup$
    – Willie Wong
    Commented Jun 12, 2020 at 2:01
  • $\begingroup$ Anyway: to your question, you maybe able to get what you want using unique continuation. $\endgroup$
    – Willie Wong
    Commented Jun 12, 2020 at 2:18
  • $\begingroup$ Thanks, I do have that n>1. $\endgroup$
    – Zamanyan
    Commented Jun 12, 2020 at 5:38

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The answer is Yes and is a consequence of Holmgren's uniqueness Theorem. See for instance Theorem 1.1.4 in this text.. The ellipticity serves here to ensure that there does not exist a characteristic hypersurface. Applying HUT, you obtain that $u\equiv0$ over the connected component of the complement of the support of $Lu$. In your case, this means $u\equiv0$ away from a finite set, but since $u\in L^\infty$, this is $u=0$.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – Zamanyan
    Commented Jun 12, 2020 at 15:48

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