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Assume that we have a differential operator such as $-\frac{\partial}{\partial x^2} + id$ on $\mathbb{R}^1$ We also then argue that if a fundamental solution has compact support, then it is supported on the origin. My follow up question is how can one then show assuming that the fundamental solution is compactly supported - that the differential operator must be of order 0?

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I myself am confused by the context in which an argument proceeds by saying that if a fundamental solution for a differential operator were compactly supported then it would have support $\{0\}$, leading to deducing that the operator is order $0$, so apparently is a multiplication operator? Or is this meant to be about pseudo-differential operators? Would you clarify the context? – paul garrett Jul 28 '12 at 19:01
I'm not sure if this includes pseudo-differential operators. The question wasn't well stated to me, I got the impression it only involved differential operators (with constant coefficients). Apparently this can be shown by an appeal to the Fourier transform of distributions... – Peadar Coyle Nov 7 '12 at 0:52
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For a constant coefficient partial differential operator P(D), the fundamental solution of P can never belong to $\epsilon'(\mathbb{R}^{n})$,i.e.have compact support.

In fact,assume we have $P(D)u=f$,where u is a distribution,then u have compact support $\Leftrightarrow$ $\frac{f}{P(\xi)}$ is analytic(The result can be found in Hormander's ALPDO, volume 1,ch7.)

Now,if we have $$P(D)u=\delta$$,obviously $\frac{1}{P(\xi)}$ is never an analytic function for a polynomia P.So the fundamental solution of P can not be compact supported.

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If the fundamental solution is supported at the origin, it must be a finite combination of derivatives of the Dirac distribution. This means that your original operator was of nonpositive order.

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