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In Friedlander's book "introduction to the theory of distributions" he claimed(on page 35):

"Now the equation $$|\langle u,\phi\rangle| \le C\sum_{|a|\le N|}\sup\{|\partial^{\alpha}\phi|:x\in K\}$$ shows that $$\langle u,\phi\rangle=0$$ if the support of $\phi$ is disjoint from $K$, so $u$ has compact support when regarded as a member of $\mathcal{D}'(X)$."

I am confused with this claim. I think in fact as a distribution $u$'s support could well be some open subset of $K$. In other words the above argument only implies

supp $u\subset K$

As the support of $u$ is defined to be the complement of the set such that functions whose support is on it vanishes when evaluated by $u$. Of course one may argue that $u$'s support is closed, and a closed subset of a compact set must be compact; but is this the correct way to interpret the claim? I feel confused.

This technical problem does not sound serious but I think without clarifying it we cannot assert that $\mathcal{E}'(X)$ can be regarded as the subspace of $\mathcal{D}'(X)$ with compact support. One direction, that any distribution with compact support can be extended to a continuous linear form is clear to me by the author's theorem. But the other direction feels not so clear.

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up vote 2 down vote accepted

A support is by definition closed, however it is defined. The definition of the support of a distribution I am familiar with, is the complement of the union of all open sets $V$ so that the distribution takes the value zero for all test functions with support inside $V$. Thus the support is indeed closed by definition.

(A partition of unity argument shows that the above-mentioned union itself has the property in question, so instead of the union we could simply say the largest open set …)

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