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 $n \in \mathbb{N}$.


1 Answer 1


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 $\langle u_{n_j},\psi_j\rangle \ne 0$ and look at $$\Psi_0 = 0, \qquad\Psi_j = \Psi_{j-1} + \frac{2^j - \langle u_{n_j},\Psi_{j-1}\rangle}{\langle u_{n_j},\psi_j \rangle} \psi_j$$ $$\Psi= \sum_j \frac{2^j - \langle u_{n_j},\Psi_{j-1}\rangle}{\langle u_{n_j},\psi_j \rangle} \psi_j \in C^\infty(\Bbb{R})$$ Then $$\lim_{j \to \infty}\langle u_{n_j},\Psi \rangle =\lim_{j \to \infty} 2^j \ \ne\ \langle u,\Psi \rangle$$ It works the same way for $\mathcal{E}'(\Omega)$ and arbitrary open $\Omega$.

  • $\begingroup$ 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? $\endgroup$
    – Math
    Aug 28, 2019 at 18:16
  • $\begingroup$ $\lim_{j \to \infty} \Psi_j$ converges because the $\psi_j$ are supported on distinct intervals. $\endgroup$
    – reuns
    Aug 28, 2019 at 18:26
  • $\begingroup$ 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. $\endgroup$
    – Math
    Aug 28, 2019 at 18:46
  • $\begingroup$ 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 $l \ge 1$ $\endgroup$
    – reuns
    Aug 28, 2019 at 18:57
  • $\begingroup$ You know what is $C^\infty_c(\Bbb{R})$ right ? The smooth functions on $\Bbb{R}$ vanishing for $|x|\ge r$. $\endgroup$
    – reuns
    Aug 28, 2019 at 19:42

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