Let $T$ be a distribution and $\varphi$ a test function.

I have read the following statement :

statement : $\text{supp}\varphi \cap \text{supp}T = \emptyset$ does not imply that $T(\varphi)=0$

I think it is wrong and I think it should be : $\varphi ^{-1} (\mathbb{C}^*)\cap \text{supp}T = \emptyset$ does not imply that $T(\varphi)=0$

I think I can prove that the initial statement it wrong, that is, I can prove that $\text{supp}\varphi \cap \text{supp}T = \emptyset \implies T(\varphi)=0$.

Reminder : let $A= \lbrace U \subset \Omega \text { such that } U \text{ open and } (\text{supp} \varphi \subset U \implies T(\varphi)=0) \rbrace$.

Then $\text{supp} T$ is the complement of $\cup_{U\in A} U$.

Proof : Let $K=\text{supp}\varphi$. By assumption, $K$ can be covered by a family of elements of $A$, and by compactness, this family can be chosen finite. Then multiplying $\varphi$ by a partition of unity $\psi$ corresponding to those elements, we get that $\varphi$ is a combination of functions supported in elements of $A$, hence by definition of $\text{supp}T$ and linearity of $T$, we get that $T(\varphi)=0$.

Am I wrong about all this ?


Your proof of $$ \big(\operatorname{supp}\varphi \cap \operatorname{supp}T = \varnothing \quad \Longrightarrow \quad T(\varphi) = 0 \big) $$ for $\varphi \in \mathcal{D}(\Omega)$ and $T \in \mathcal{D}'(\Omega)$ is correct.

For $T \in \mathcal{D}'(\Omega)$ and an open set $U \subseteq \Omega$ one defines the restriction $T|_{U}\in \mathcal{D}'(U)$ of $T$ to $U$ by $$ T|_{U}(\varphi) := T(\varphi) $$ for $\varphi \in \mathcal{D}(U)$. Now one naturally expects that $$ T|_{\Omega \setminus \operatorname{supp}T}=0 $$ and you proved that $T(\varphi)=0$ for $\varphi$ with $\operatorname{supp}\varphi \subseteq \Omega \setminus \operatorname{supp}T$.

Somehow related: Distributions have Sheaf structure.


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