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

  • $\begingroup$ 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 '21 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 '21 at 11:01
  • $\begingroup$ Do you think this is the only way to construct a continuous extension of $u$? $\endgroup$
    – Eric
    Aug 28 '21 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 '21 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 '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 $L$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.