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Suppose we have a distribution $F \in \mathcal{D}'(\mathbb{R})$ and we know that for every $\phi \in \mathcal{D}(\mathbb{R})$ with $\text{supp}\phi \subseteq \text{supp}F$, we have $$ (F, \phi) = 0 .$$ Does this imply that $(F, \phi) = 0 $ for all test functions $\phi \in \mathcal{D}(\mathbb{R})$?

If we have a test function with support which intersects with supp$F$ and also (supp$F)^{c}$ then I am not sure how to show it in that case. My first idea would have been to split $\phi$ up into two parts, i.e. $\phi = \phi \mathbf{1}_{suppF} + \phi \mathbf{1}_{(suppF)^{c}}$ , but then the two parts would not necessarily be test functions anymore so that doesn't work I think

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Let $F(\phi)=\phi(0)$, [the delta disribution]. Then $supp (F)=\{0\}$. If $\phi \in \mathcal D(\mathbb R)$ and $supp (\phi) \subseteq \{0\}$ then $\phi \equiv 0$ (by continuity) so $F(\phi)=0$. On the other hand there exist functions $\psi \in D(\mathbb R)$ with $F(\psi)=\psi (0) \neq 0$.

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  • $\begingroup$ Of course thank you, do you think there can be an example where $suppF$ is a positive measure set? $\endgroup$
    – duelspace
    Commented Nov 12, 2023 at 11:38
  • $\begingroup$ Yes, you can take a measure whose support is a fat Cantor set. $\endgroup$ Commented Nov 14, 2023 at 13:45
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    $\begingroup$ @LorenzoPompili I wanted to say Cantor set of positive measure. This works because such a set has no interior. $\endgroup$ Commented Nov 14, 2023 at 13:47
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A counterexample where the support of $F$ has positive measure. Take a closed set $A$ with positive measure and empty interior (edit: as @geetha290krm suggested, you can take an $\varepsilon$-Cantor set), and take $F=\chi_A$ the indicator function of $A$, where we identify $L^1_{\operatorname{loc}}(\mathbb R)$ as a subspace of $\mathscr D’(\mathbb R)$ in the known way. As a side note, the support of $A$ can be chosen so that the complement $A^c$ has arbitrarily small measure.

Since $A$ is closed, it is immediate that $\operatorname{supp (F)}\subseteq A$. Moreover, $\operatorname{supp}(F)$, i.e., the support of $F$ as a distribution, coincides with the essential support of $\chi_A$ as an $L^1_{\operatorname{loc}}$ function (this is a general fact for functions in $L^1_{\operatorname{loc}}$, which is straightforward to prove, I did not find a reference though). In particular, $A\setminus \operatorname {supp} (F)$ has zero measure, thus $\operatorname {supp} (F)$ has strictly positive measure.

The above construction yields a counterexample: any test function supported on $\operatorname{supp}(F)$ must be identically zero, because the complement of $A$ (hence that of $\operatorname{supp}(F)$) is dense in $\mathbb R$, and the support of any non-trivial test function has non-empty interior.

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