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Let $p$ be a polynomial such that $p(D)$ is a hypoelliptic operator, i.e. if $u$ is a distribution satisfying $p(D)u = 0$ then $u$ is smooth. Here, $D = -i\partial$.

Let $E$ be a fundamental solution for $p(D)$; that is, assume that $E$ is a distribution such that $p(D)E = \delta$ where $\delta$ is the Dirac distribution.

I have read in a few texts and resources that since $p(D)E = 0$ away from the origin, $E$ must be smooth away from $0$. However, in every reference I've found, this has been stated without any kind of proof or justification. Nonetheless, this is not at all clear to me.

To prove the statement more rigorously, I tried multiplying $E$ by cut-off functions but that idea doesn't seem to lead anywhere. How could I go about proving the result?

EDIT: I know by assumption that if $u$ satisfies $p(D)=0$ everywhere then it is smooth. If $E$ is a fundamental solution for $p(D)$ then $p(D)E = 0$ away from the origin. So if I can show that p(D) is "local" I would be done. By this, I meant that I would like to show the following;

If $p(D)u = 0$ in some neighbourhood, then $u$ is smooth in that neighbourhood.

When I tried to introduce a cut-off function $\zeta$ as in the comments, I was still only able to obtain $p(D)(E\zeta) = 0$ outside of some small set - but not everywhere. So I was not able to directly apply my hypothesis.

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  • $\begingroup$ I think that you also need to use that $p(D)$ is a local operator, that is, $p(D)f(x)$ only depends on $f$ in a neighborhood of $x$. $\endgroup$ – Giuseppe Negro Mar 27 at 0:41
  • $\begingroup$ @GiuseppeNegro I don't know enough to agree or disagree with you. However, I did see the statement without the assumption that p(D) is a local operator in my course's lecture notes $\endgroup$ – Quoka Mar 27 at 0:49
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    $\begingroup$ I mean, fix a point x different from the origin, and try to prove that E is smooth in a neighborhood of x, by multiplying it with a smooth function that equals 1 in a neighborhood of x and whose support does not contain 0. $\endgroup$ – Giuseppe Negro Mar 27 at 1:09
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    $\begingroup$ @GiuseppeNegro That was what I tried to do. So suppose $\zeta$ is a smooth function such that $\zeta= 1$ near $x$ and $\zeta = 0$ outside of a sufficiently small ball about $x$. Then $p(D)(\zeta E) = \zeta p(D)E + v = \zeta\delta + v = v$. So if I can show that $v\equiv 0$ then I know that $\zeta E$ is smooth. In particular, $E$ is smooth at $x$. But I didn't manage to gather information on $v$. $\endgroup$ – Quoka Mar 27 at 1:13
  • $\begingroup$ @Quoka Isn't it $v = Ep(D)\zeta \equiv 0$ in a neighborhood of x? $\endgroup$ – ares Apr 5 at 2:18

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