Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 …)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.