# Smoothness of fundamental solution of a hypoelliptic operator.

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.

• 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 $x$. – Giuseppe Negro Mar 27 at 0:41
• @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 – Quoka Mar 27 at 0:49
• 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. – Giuseppe Negro Mar 27 at 1:09
• @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 $v$. – Quoka Mar 27 at 1:13
• @Quoka Isn't it $v = Ep(D)\zeta \equiv 0$ in a neighborhood of x? – ares Apr 5 at 2:18