On the relation between $\scr D'$ and $\scr E'$

Let $$u \in {\scr D}'(\Omega)$$, with $$\operatorname{supp} u =K$$ compact. Then $$\exists! \ \tilde u \in {\scr E}'(\Omega)$$ such that $$\tilde u (f) = u(f)$$ holds $$\forall f \in {\scr D}(\Omega)$$. In other terms:

Every compactly supported distribution can be uniquely associated to a distribution in $$\scr E'(\Omega)$$

But... what can be said vice versa? Are there distributions in $${\scr E'}$$ which cannot be seen as compactly supported distributions in $$\scr D'$$?

Addendum. Let's specify some definitions for clarity.

$${\scr D}(\Omega)$$ denotes the set of all compactly supported smooth functions, $$f \in {\cal C}^\infty_0(\Omega)$$, equipped with the following convergence notion:

$${\cal C}^\infty_0(\Omega) \ni \{ f_n \}_{n \in \mathbb N} \overset {n \to \infty}{\longrightarrow} f \in {\scr C}^\infty_0(\Omega)$$ if:

• $$\exists K \subset \Omega$$ compact $$\ |$$ $${\rm supp}(f_n) \subseteq K \quad \forall \ n \$$ and
• $$\partial^\alpha f_n \rightrightarrows \partial^\alpha f$$ uniformly, $$\forall \alpha$$ multiindex.

A linear functional $$u: {\scr D}(\Omega) \to \mathbb C$$ is a distribution if one of this two equivalent conditions applies:

1. $$\forall$$ convergent sequence $$\{f_n\}_{n \in \mathbb N} \in {\cal D}(\Omega)$$ one has: $$\lim_n u(f_n) = u (\lim_n f_n)$$

2. $$\forall K \subset \Omega \$$ compact$$\ \exists c_K > 0, N_K \in \mathbb N \cup \{0\} \$$ such that $$\ \forall f$$ with $$\operatorname{supp} f \subseteq K$$ $$|u(f)| \leq c_K \displaystyle \sum_{|\alpha| \leq N_K} \underset {K}{\sup} |\partial^\alpha f|$$

Lastly, $${\scr D}' (\Omega) \equiv \{ u: {\scr D}(\Omega) \to \mathbb C \ | \ u$$ is a distribution$$\}$$

$${\scr E}(\Omega)$$ denotes the set of all smooth functions, $$f \in {\cal C}^\infty(\Omega)$$, equipped with the following convergence notion:

$${\cal C}^\infty(\Omega) \ni \{ f_n \}_{n \in \mathbb N} \overset {n \to \infty}{\longrightarrow} f \in {\cal C}^\infty(\Omega)$$ if

• $$\partial^\alpha f_n \rightrightarrows \partial^\alpha f$$ uniformly, $$\forall \alpha$$ multiindex, and $$\forall K \subset \Omega$$ compact.

A linear functional $$u: {\scr E}(\Omega) \to \mathbb C$$ is a distribution if one of this two equivalent conditions applies:

1. $$\forall$$ convergent sequence $$\{f_n\}_{n \in \mathbb N} \in {\cal E}(\Omega)$$ one has: $$\lim_n u(f_n) = u (\lim_n f_n)$$

2. $$\exists K \subset \Omega \$$ compact$$, c_K > 0, N_K \in \mathbb N \cup \{0\} \$$ such that $$\ \forall f \in {\scr E}(\Omega)$$ $$|u(f)| \leq c_K \displaystyle \sum_{|\alpha| \leq N_K} \underset {K}{\sup} |\partial^\alpha f|$$

Lastly, $${\scr E}' (\Omega) \equiv \{ u: {\scr E}(\Omega) \to \mathbb C \ | \ u$$ is a distribution$$\}$$

The point is it's not at all obvious if $${\scr E}'$$ consists of all and only those compactly supported distributions, or something else, from these definitions...

• How is $\mathscr{E}'(\Omega)$ defined here? When I studied distribution theory it was defined as the set of compactly supported distributions. Jul 5 '21 at 19:13

Let $$\mathcal{D}'_c(\Omega)$$ denote compactly supported distributions in $$\mathcal{D}'(\Omega).$$

Theorem: If $$u\in\mathcal{D}'(\Omega)$$ has compact support, then we can extend its domain to $$\mathcal{E}(\Omega)$$.

Proof: Take $$\rho\in \mathcal{D}(\Omega)$$ such that $$\rho\equiv 1$$ on a neighborhood of $$\operatorname{supp}u.$$ Then, for $$\varphi\in \mathcal{E}(\Omega),$$ set $$\langle u, \varphi \rangle := \langle u, \rho\varphi \rangle.$$ It is clear that $$\rho\varphi \in \mathcal{D}(\Omega)$$ so the last expression is defined. The definition is also not dependent of the choice of $$\rho$$ since two such choices ($$\rho_1,\rho_2$$) only differ outside of $$\operatorname{supp}u$$ making $$\langle u, (\rho_1-\rho_2)\varphi\rangle = 0.$$

This makes $$\mathcal{D}'_c(\Omega) \subseteq \mathcal{E}'(\Omega).$$

Theorem: If $$u\in\mathcal{E}'(\Omega)$$ then $$u$$ has compact support.

Idea of Proof: Assume that $$u\in\mathcal{E}'(\Omega)$$ has not compact support. Then there is an infinite number of disjoint compact sets $$K_n$$ on which $$u\neq 0.$$ For each such $$K_n$$ choose $$\varphi_n\in\mathcal{D}(K)$$ such that $$\langle u, \varphi_n \rangle = 1.$$ Then $$\sum_n \varphi_n \in \mathcal{E}(\Omega)$$ so $$\langle u, \sum_n \varphi_n \rangle$$ is finite. But $$\langle u, \sum_n \varphi_n \rangle = \sum_n \langle u, \varphi_n \rangle = \sum_n 1 = \infty.$$ Contradiction!

This makes $$\mathcal{E}'(\Omega) \subseteq \mathcal{D}'_c(\Omega).$$

Thus, $$\mathcal{E}'(\Omega) = \mathcal{D}'_c(\Omega).$$

• Would you mind helping me with the question: In the proof of the first theorem, you mention that the extension of $u$ is independent of choice of cut-off functions; yet, it seems to me that it will be sufficient if we already set $\langle u, \varphi \rangle = \langle u, \rho \varphi \rangle$ since it is well-defined; I mean it seems like we don't necessarily point out it makes no difference if we choose different cut-off function to prove the definition is well-defined. Am I right?
– Eric
Aug 27 '21 at 10:21
• @Eric. I think you're right. One choice of $\rho$ gives a well-defined extension $u_\rho$. What the independence of choice shows is uniqueness. Aug 27 '21 at 11:01
• Do you think this is the only way to construct a continuous extension of $u$?
– Eric
Aug 28 '21 at 9:40
• @Eric. I think so. Let $\hat{u}$ be an extension, and take some $\rho$ as above. Then $\hat{u} = \rho\hat{u} + (1-\rho)\hat{u},$ where the first term equals $u$ and the second term has support only where $u\equiv 0$ and therefore also should vanish. Aug 28 '21 at 12:02
• So...it suggests that $\left(1-\rho\right)\hat{u}\equiv 0$ (on $C^{\infty}\left(\Omega\right)$) and note that, by definition, $\rho\hat{u}\left(\varphi\right)=\hat{u}\left(\rho \varphi\right)=\langle u,\varphi\rangle$ on $C^{\infty}\left(\Omega\right)$. Hence, we can see any continuous extension $\hat{u}$ equals to $\langle u,\varphi\rangle$, as constructed by cut-off functions. Right?
– Eric
Aug 29 '21 at 7:38

As in @md2perpe's comment: this partly depends on the choice of logical order in the exposition.

E.g., we can either define $$\mathcal E'$$ as compactly-supported distributions, or as the continuous dual of the space $$\mathcal E=C^\infty$$ of smooth functions, and then prove that it consists of compactly-supported distributions, using the natural inclusion of test functions to smooth.

But/and, yes, a compactly supported distribution can be defined as a continuous linear functional on smooth functions, via smooth cut-offs (and showing well-definedness).

So, in any case, whatever the logical order, there is an exact identification (by various reasonable means) of $$\mathcal E'$$ with compactly supported distributions.

The reason that every continuous linear functional $$u$$ on $$\mathscr E(\Omega)$$ has compact support is the continuity estimate for $$u$$: Recall that the topology of $$\mathscr E(\Omega)$$ is defined by the seminorms $$\|f\|_{K,n}=\sup\{|\partial^\alpha f(x)|: x\in K, |\alpha|\le n\},$$ and hence the are a compact set $$K\subseteq\Omega$$, a differentiability order $$n$$ and a constant $$c$$ such that $$|u(f)|\le c\|f\|_{K,n}$$ for all $$f\in\mathscr E(\Omega)$$. If $$L$$ is another compact set containing $$K$$ in its interior you then get (using a cut-off function) that $$u$$ (considered as an element of $$\mathscr D'(\Omega)$$) has support in $$L$$.