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...

  • $\begingroup$ How is $\mathscr{E}'(\Omega)$ defined here? When I studied distribution theory it was defined as the set of compactly supported distributions. $\endgroup$
    – md2perpe
    Jul 5, 2021 at 19:13

3 Answers 3


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

  • $\begingroup$ 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? $\endgroup$
    – Eric
    Aug 27, 2021 at 10:21
  • 1
    $\begingroup$ @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. $\endgroup$
    – md2perpe
    Aug 27, 2021 at 11:01
  • $\begingroup$ Do you think this is the only way to construct a continuous extension of $u$? $\endgroup$
    – Eric
    Aug 28, 2021 at 9:40
  • 1
    $\begingroup$ @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. $\endgroup$
    – md2perpe
    Aug 28, 2021 at 12:02
  • $\begingroup$ 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? $\endgroup$
    – Eric
    Aug 29, 2021 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 $L$.


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