# A convergent sequence in $\mathcal{E}'(\Omega)$ is supported in a single compact set

Let $$\mathcal{E}'(\Omega)$$ be the space of the distributions with compact support. Suppose that $$u_n \rightarrow u$$ in $$\mathcal{E}'(\Omega)$$. I'm trying to show that there exist a compact $$K\subset \Omega$$ such that $$\operatorname{supp} u_n, \operatorname{supp} u \subset K$$ for all $$n \in \mathbb{N}$$.

My attempt: Let us show that there exist $$n_0 \in \mathbb{N}$$ and a compact $$K_1$$ such that $$\operatorname{supp} u_n, \operatorname{supp} u \subset K_1$$ for all $$n \geq n_0$$. Then setting $$K_2=\bigcup_{n=1}^{n_0-1} \operatorname{supp} u_n$$ we obtain $$K=K_1\cup K_2$$ which is a compact subset set of $$\Omega$$ and $$\operatorname{supp} u_n, \operatorname{supp} u \subset K$$ for all $$n \in \mathbb{N}$$.

Indeed, since $$u_n \rightarrow u$$ in $$\mathcal{E}'(\Omega)$$, then $$\langle u_n, \varphi \rangle_{\mathcal{E}'(\Omega), \mathcal{E}(\Omega)} \rightarrow \langle u, \varphi \rangle_{\mathcal{E}'(\Omega), \mathcal{E}(\Omega)} \hbox{ for all } \varphi \in \mathcal{E}(\Omega)=C^\infty(\Omega).$$ Let $$K_1=\operatorname{supp} u$$ and $$\varphi \in C_0^{\infty}(\Omega)$$ such that $$\operatorname{supp} \varphi \subset \Omega \setminus K_1$$. Thus, for all $$\varepsilon>0$$ there exist $$n_0(\varphi) \in \mathbb{N}$$ such that $$|\langle u_n, \varphi \rangle_{\mathcal{E}'(\Omega), \mathcal{E}(\Omega)} -\langle u, \varphi \rangle_{\mathcal{E}'(\Omega), \mathcal{E}(\Omega)}|<\varepsilon \hbox{ for all } n \geq n_0.$$ Since $$\operatorname{supp} u \cap \operatorname{supp} \varphi =\emptyset$$, then $$\langle u, \varphi \rangle_{\mathcal{E}'(\Omega), \mathcal{E}(\Omega)}=0.$$

Therefore, for all $$\varepsilon>0$$ there exist $$n_0(\varphi) \in \mathbb{N}$$ such that $$|\langle u_n, \varphi \rangle_{\mathcal{E}'(\Omega), \mathcal{E}(\Omega)}|<\varepsilon \hbox{ for all } n \geq n_0.$$ From the arbitrariness of $$\varepsilon>0$$ we obtain $$\langle u_n, \varphi \rangle_{\mathcal{E}'(\Omega), \mathcal{E}(\Omega)}=0 \hbox{ for all } n \geq n_0.$$ Since the function $$\varphi$$ was taken arbitrarily, we have $$U=\Omega \setminus K_1$$ is a open subset of $$\Omega$$ such that $$u_n=0 \hbox{ in } U \hbox{ for all } n \geq n_0.$$

(I'm not sure about this part, because the constant $$n_0$$ depends on $$\varphi$$).

Therefore, $$\operatorname{supp} u_n \subset \Omega \setminus U=K_1 \hbox{ for all } n \geq n_0.$$

Defining $$K_2=\bigcup_{n=1}^{n_0-1} \operatorname{supp} u_n$$ and $$K=K_1\cup K_2$$, it's easy to see that $$K$$ is a compact subset set of $$\Omega$$ and $$\operatorname{supp} u_, \operatorname{supp} u \subset K$$ for all $$n \in \mathbb{N}$$.

Say the $$u_n$$ are compactly supported distributions $$\in D'(\Bbb{R})$$.
If they are not all supported on a common compact, looking at the smallest interval containing $$supp(u_n)$$ find $$n_j \to \infty,|k_j|>|k_{j-1}| +2$$ such that $$u_{n_j}$$ is zero on $$|x| >|k_j|+1$$ but not on $$(k_j,k_j+1)$$,
Take $$\psi_j\in C^\infty_c(k_j,k_j+1)$$ such that $$\ne 0$$ and look at $$\Psi_0 = 0, \qquad\Psi_j = \Psi_{j-1} + \frac{2^j - }{} \psi_j$$ $$\Psi= \sum_j \frac{2^j - }{} \psi_j \in C^\infty(\Bbb{R})$$ Then $$\lim_{j \to \infty}=\lim_{j \to \infty} 2^j \ \ne\ $$ It works the same way for $$\mathcal{E}'(\Omega)$$ and arbitrary open $$\Omega$$.
• I did not understand the construction of the compacts and why the sequence $\Psi_j$ converges in $C^\infty(\mathbb{R})$. Could you give more details? – Victor Hugo Aug 28 at 18:16
• $\lim_{j \to \infty} \Psi_j$ converges because the $\psi_j$ are supported on distinct intervals. – reuns Aug 28 at 18:26
• Sorry, the proof is very complicated for me. Could you explain why $\Psi= \sum_j \frac{2^j - <u_{n_j},\Psi_{j-1}>}{<u_{n_j},\psi_j>} \psi_j$ converges? I also don't know how to guarantee that $\langle u_{n_j}, \Psi \rangle= 2^j$, since we have a double limit. – Victor Hugo Aug 28 at 18:46
• The $\psi_j$ are smooth functions supported on disjoint intervals. On an interval $\Psi$ is given by finitely many $\psi_j$. That we obtain $<u_{n,j},\Psi>=2^j$ is because $<u_{n_j},\psi_{j+l}>=0$ for $l \ge 1$ – reuns Aug 28 at 18:57
• You know what is $C^\infty_c(\Bbb{R})$ right ? The smooth functions on $\Bbb{R}$ vanishing for $|x|\ge r$. – reuns Aug 28 at 19:42