> [!theorem] > > Let $U \subset \real^d$ be an [[Open Set|open set]] and $\seqi{V}$ be an [[Open Cover|open cover]] of $U$. Suppose that > 1. For each $i \in I$, there exists a [[Distribution|distribution]] $T_i \in \cd'(V_i)$. > 2. For any $i, j \in I$ with $V_i \cap V_j \ne \emptyset$, $T_i|_{\cd(V_i \cap V_j)} = T_j|_{\cd(V_i \cap V_j)}$. > > then there exists a unique distribution $T \in \cd'(U)$ such that $T|_{\cd(V_i)} = T_i$ for all $i \in I$. > > *Proof*. Let $\phi \in \cd(U)$, $\bracs{V_{i_m}}_1^M \subset \seqi{V}$ finite subcover of $\supp{\phi}$. Let $\bracs{\psi_m}_1^M$ be a [[Extension Lemma for Smooth Functions|smooth]] [[Partition of Unity|partition of unity]] subordinate to $\bracs{V_{i_m}}_1^M$, and define > $ > \angles{T, \phi} = \sum_{m = 1}^M \angles{T_{i_m}, \psi_m \phi} > $ > Suppose that $\bracs{V_{j_n}}_1^N \subset \seqi{V}$ is another finite subcover, $\bracs{\chi_n}_1^N$ is a smooth partition of unity subordinate to it, then > $ > \begin{align*} > \sum_{m = 1}^M \angles{T_{i_m}, \psi_m \phi} &= \sum_{m = 1}^M \sum_{n = 1}^N\angles{T_{i_m}, \psi_m\chi_n \phi} \\ > &= \sum_{n = 1}^N\sum_{m = 1}^M \angles{T_{j_n}, \psi_m\chi_n \phi}= \sum_{n = 1}^N\angles{T_{j_n}, \chi_n \phi} > \end{align*} > $ > so $T$ is well-defined. Now let $\seq{\phi_n} \subset \cd(U)$ such that $\phi_n \to \phi$ in $\cd(U)$. Let $K \subset U$ be a [[Compactness|compact]] set such that $\supp{\phi_n} \subset K$ for all $n \in \nat$. Let $\bracs{V_{i_m}}_1^M \subset \seqi{V}$ be a finite subcover of $K$ and $\bracs{\psi_m}_1^M$ be a partition of unity subordinate to it, then > $ > \angles{T, \phi_n} = \sum_{m = 1}^M \angles{T_m, \psi_m \phi_n} \to \sum_{m = 1}^M \angles{T_m, \psi_m \phi} = \angles{T, \phi} > $ > as $n \to \infty$, so $T$ is a distribution. > [!theorem] > > Let $U \subset \real^d$ be an [[Open Set|open set]] and $\seqi{V}$ be an [[Open Cover|open cover]] of $U$. Let $\seq{S_n} \subset \cd'(U)$ be a sequence of distributions such that: > 1. For each $i \in I$, there exists a [[Distribution|distribution]] $T_i \in \cd'(V_i)$ such that $S_n|_{\cd(V_i)} \to T_i$ in $\cd'$. > 2. For any $i, j \in I$ with $V_i \cap V_j \ne \emptyset$, $T_i|_{\cd(V_i \cap V_j)} = T_j|_{\cd(V_i \cap V_j)}$. > > then > 3. There exists a unique $T \in \cd'(U)$ such that $T|_{\cd(V_i)} = T_i$ for all $i \in I$. > 4. $S_n \to T$ in $\cd'$. > > *Proof*. From the gluing lemma, result $(1)$ holds directly. Now let $\phi \in \cd(U)$, $\bracs{V_{i_m}}_1^M$ be a finite subcover of $\supp \phi$, $\bracs{\psi_{i_m}}_1^M$ be a partition of unity subordinate to it, then > $ > \angles{S_n, \phi} = \sum_{m = 1}^M \angles{S_n, \psi_{i_m}\phi} \to \sum_{m = 1}^M \angles{T_{i_m}, \psi_{i_m}\phi} = \sum_{m = 1}^M \angles{T, \psi_{i_m}\phi} = \angles{T, \phi} > $ > so $S_n \to T$ in $\cd'$.