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First some definitions. $\Omega \subseteq \mathbb R^N$.

  1. ${\cal D}(\Omega)$ 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 {\cal C}^\infty_0(\Omega) $ if:

  • $\exists K \subset \Omega$ compact $\ |$ ${\rm supp}(f_n) \subseteq K \quad \forall n$
  • $\partial^\alpha f_n \rightrightarrows \partial^\alpha f $ uniformly, $\forall \alpha $ multiindex.
  1. a linear functional $u: {\cal D}(\Omega) \to \mathbb C $ is a distribution if $\forall$ convergent sequence $ \{f_n\}_{n \in \mathbb N} \in {\cal D}(\Omega)$ one has: $$\lim_n u(f_n) = u (\lim_n f_n)$$

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

  1. given $\Omega_0 \subset \Omega $, $\Omega_0$ is said to be a vanishing open set for $u \in {\cal D}'(\Omega) $ if $\forall f \in {\cal D}(\Omega) $ such that ${\rm supp} f \subseteq \Omega_0$, one has: $$u(f) =0$$

4.$ \forall u \in {\cal D}'(\Omega) $ we define ${\rm supp } \ u \equiv \Omega \setminus \bigcup\limits^{}_{i} \Omega_0^i$, with $\Omega_0^i$ a vanishing open set for $u$.


Question: is ${\rm supp} \ u $ compact $\forall u \in {\cal D}'(\Omega) $?

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  • $\begingroup$ Since $\Omega \subseteq \mathbb R^N$, which is finite-dimensional, maybe we could apply Heine-Borel theorem: $\Omega$ is compact $\iff$ $\Omega$ is closed and bounded. It's closed by definition, as a complementary set of a union of open sets which is open, but I'm not sure about boundedness. $\endgroup$
    – ric.san
    Jun 27, 2021 at 14:28

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No, there are distributions with non-compact support. An example is a pulse train: $$ u = \sum_{k=-\infty}^{\infty} \delta_k, $$ where $\delta_k$ is a Dirac delta supported at $x=k$. Less exotic examples come from non-compactly supported functions in $L^1_{loc}$.

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  • $\begingroup$ Is ${\rm supp } \ u \equiv \mathbb Z$ in this case? $\endgroup$
    – ric.san
    Jun 27, 2021 at 14:41
  • $\begingroup$ Yes, that’s correct. $\endgroup$
    – mrf
    Jun 27, 2021 at 14:41
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    $\begingroup$ Another example is $u(f):=\int_{-\infty}^{\infty} f(x) \, dx.$ It has all of $\mathbb{R}$ as support. $\endgroup$
    – md2perpe
    Jun 27, 2021 at 17:03

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