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In my book I have the following definition: Let $X\subset\mathbb{R^{n}}$ be an open set, and let $u\in \mathcal{D}'(X)$. The support of $u$ written as supp $u$, is the complement of the set

$$\{x : u=0 \quad \mbox{on a neighbourhood of}\, x\}.$$ There is also a remark, that it is a cloed subset of $X$.

My first question is: what does that mean that distribution vanishes, i.e., $u=0$ and $u=0$ on a neighbourhood of $x$?

I was thinking as follows: consider all $\varphi\in C_c^{\infty}(X)$ such that $$<u,\varphi>=0.\qquad (1)$$ Every function satisfying (1) has a compact support, namely $K_{\varphi}$. Now, for every $K_{\varphi}$ I can pick a "cut-off" function $\psi_{\varphi}$ which equales $1$ in the neighbourhood of $K_{\varphi}$ and $0$ elsewhere. Let $A_{\varphi}$ be a set of all poinsts such that $\psi_{\varphi}=0$ excluding the points from outside the support $K_{}\varphi$. Let $A=\bigcup A_{\varphi}$. Finally, consider the complement of $A$. Can take it for the definition of the support?

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If $u \in \mathcal{D}'(\Omega)$, then $u=0$ in the largest open $G$ of $\Omega$. In the sense that if $G$ is the union of all the open sets of $\Omega$ such that $u=0$, then for each fixed $\varphi \in \mathcal{D}_K(G)$, let $K \subset \Omega_1 \cup \cdot \cdot \cdot \cup \Omega_m$, that is $u=0$ on $\Omega_j$ and consider a regular unit partition $\lbrace \varphi_j \rbrace_{j=1}^{m}$ with $\mathrm{supp}(\varphi_j) \subset \Omega_j$ (subordinate to this covering of $K$), in the sense of the partition of unit we have \begin{align*} u(\varphi)=u(\varphi \sum_{j=1}^{m} \varphi_j)=\sum_{j=1}^{m} u(\varphi \varphi_j)=0 \end{align*} since $\sum_{j=1}^m \varphi_j=1$ and $\mathrm{supp}(\varphi \varphi_j) \subseteq \mathrm{supp}(\varphi_j) \subset \Omega_j$ . Now the support of $u \in \mathcal{D}'(\Omega)$, $\mathrm{supp}(u)$ is defined as complementary $\Omega \setminus G$ of the largest open subset $G$ of $\Omega$ where $u$ is zero, i.e. \begin{align*} \mathrm{supp}(u) = \lbrace x \in \Omega : \nexists U_x \subset \Omega : u_{U_x}=0 \rbrace \end{align*}

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