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Let $\phi\in D(\Omega)$ and $f\in D'(\Omega)$. If $\phi$ is $0$ in a neighbourhood of $\operatorname{supp}(f)$, then how will we prove that $\langle f, \phi \rangle$ is also $0$? Will it be sufficient if $\phi$ vanishes on $\operatorname{supp}(f)$?

Definition of $\operatorname{supp}(f)$:

Let $f$ be a distribution, and $U$ an open set in $\Bbb R^n$ such that, for all test functions $\phi$ with the support of $\phi$ contained in $U$, $f(\phi) = 0$. Then $f$ is said to vanish on $U$. Hence we can define the support of $f$ as the complement of the largest open set on which $f$ vanishes.

Alternatively, we can say $\operatorname{supp}(f)$ is the subset of $\Bbb R^n$ such that $x \in \operatorname{supp}(f)$, if and only if for all neighbourhoods $\mathcal U$ of $x$, there is $\phi \in D(\mathcal U)$ such that $\langle f, \phi\rangle \neq 0$.

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    $\begingroup$ Yes. This follows directly from the definition of a distributions support. $\endgroup$
    – goonfiend
    May 27, 2013 at 4:23
  • $\begingroup$ How? @SeanGomes $\endgroup$ May 27, 2013 at 4:27
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    $\begingroup$ @user76097 I don't know if the bold part is a question. However, this is not sufficient, you need $\phi$ vanishing on a neighbourhood of supp $f$. $\endgroup$
    – Vobo
    May 27, 2013 at 6:06
  • $\begingroup$ @Vobo yeah it is the part of this question. $\endgroup$ May 27, 2013 at 11:58

2 Answers 2

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If $\phi$ vanishes only on $\operatorname{supp } f$, then this is not true anymore: Let $f=\delta'$ and and $\phi\in C_c(\mathbb{R})$ with $\phi(x)=x$ for $|x|\leq 1$. Then $\phi(0)=0$, but $f(\phi)= -\phi'(0) = -1$.

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  • $\begingroup$ @Vovo thanks a lot. I got your point. You gave a very good example. $\endgroup$ May 27, 2013 at 13:35
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If $\phi$ is $0$ in a neighbourhood of $\textrm{supp}(f)$, then the support of $\phi$ is a closed set that does not meet $\textrm{supp}(f)$. In particular this implies that $f$ vanishes on $\textrm{supp}(\phi)$ in the sense of your definition. So $f(\phi)=0.$

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