# Distribution vanishes outside the support?

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

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 $$F(\psi)=\psi (0) \neq 0$$.
• Of course thank you, do you think there can be an example where $suppF$ is a positive measure set? Commented Nov 12, 2023 at 11:38
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.