I am reading several books about distribution theory (or generalized function theory). For a distribution $u \in \mathcal{D}'(\mathbb{R}^n),$ some books define its support set as follows:
$\textbf{Definition1:}$ Support of $u$ is intersection of all closed set $K$ satisfying $(u,\phi)=0$ for all $\phi \in \mathcal{D}(\mathbb{R}^n)$ and $\text{supp} \phi \in K^c.$
On the other hand, for a continuous function $u \in C(\mathbb{R}^n),$ the classical definition of its support is
$\textbf{Definition2:}$ Support of $u:=\overline{\{x \in \mathbb{R}^n:u(x)\neq 0\}}$.
I am curious that are the two definitions equivalent when $u$ is a distribution and also a continuous function?
For certain $x$ in the support set of $u$ under Def 1, then $x$ is in the intersection of all the closed set $K$ satisfying $\int u\phi \ dx=0$ for any $\phi \in \mathcal{D}(K^c),$ thanks to the continuity of $u,$ we have $u=0$ on $K^c.$ Then $x$ is in the intersection of all the closed set $K$ where $u$ is zero on the open set $K^c.$ Thus I need to show that $$ \cap_{K}\{K:K \ \text{is closed,}u|_{K^c}=0\}=\overline{\{x \in \mathbb{R}^n:u(x)\neq 0\}}. $$
Clearly, $$ \cap_{K}\{K:K \ \text{is closed,}u|_{K^c}=0\} \subset \overline{\{x \in \mathbb{R}^n:u(x)\neq 0\}}, $$ since $\overline{\{x \in \mathbb{R}^n:u(x)\neq 0\}}$ is a closed set where u vanishes on its complement.(Is it true? To prove the bold sentence, should I need to use the fact that $u$ is continuous?)
On the other hand, for every closed set K where u vanishes on its complement, we have $$ \overline{\{x \in \mathbb{R}^n:u(x)\neq 0\}} \subset K. $$ Therefore we conclude that $$ \overline{\{x \in \mathbb{R}^n:u(x)\neq 0\}} \subset \cap_{K}\{K:K \ \text{is closed,}u|_{K^c}=0\}. $$
My question is the above deduction is right? And how to clearify the bold sentence in my deduction? Can anyone give me some advice?