I am aware that the dual of $C^\infty(\mathbb{R}^n)$ is the space of distributions (not necessarily tempered) with compact support.


However, if we fix a compact set $K \subset \mathbb{R}^n$, is the space of distributions supported on $K$ the dual of some test function space?


If so, the test function space must be even larger than $C^\infty(\mathbb{R}^n)$, so I am a bit confused...


Could anyone please clarify?



Given a compact set $K \subset \mathbb{R}^n$, is the space of distributions supported on $K$ the dual of some test function space?
A few reminders:
[1] The dual space of $\mathscr D(\mathbb R^n)=
C^\infty\_c(\mathbb R^n)$ ($C^\infty$ functions with compact support) is
$\mathscr D'(\mathbb R^n)$ (distributions on $\mathbb R^n$).


[2] The dual space of $\mathscr S(\mathbb R^n)$
($C^\infty$ functions rapidly decreasing) is
$\mathscr S'(\mathbb R^n)$ (temperate distributions on $\mathbb R^n$).


[3] The dual space of $\mathscr E(\mathbb R^n)=C^\infty(\mathbb R^n)$
($C^\infty$ functions) is
$\mathscr E'(\mathbb R^n)$ (compactly supported distributions on $\mathbb R^n$).


You have the continuous injections
$
\mathscr D(\mathbb R^n)\subset \mathscr S(\mathbb R^n)\subset \mathscr E(\mathbb R^n)
$
and thus
$$
\mathscr D'(\mathbb R^n)\supset \mathscr S'(\mathbb R^n)\supset \mathscr E'(\mathbb R^n).
$$
The only thing which is not obvious in the above statements is [3]: you have to prove that a continuous linear form on $\mathscr E(\mathbb R^n)$ has compact support. It is possible to prove that with the Paley-Wiener Theorem.


Now if you fix a compact set $K$ and consider $\mathscr D\_K(\mathbb R^n)=C\_K^\infty(\mathbb R^n)$, then the latter space is a Fréchet space (which is complete and metrizable); note that $\mathscr D(\mathbb R^n)$ is the inductive limit of the $\mathscr D\_K(\mathbb R^n)$ and is not metrizable.
If you want to consider $\mathscr E'\_K(\mathbb R^n)$ (the distributions supported in $K$, assumed to be convex), then you have a characterization via the Paley-Wiener theorem: $u$ belongs to
$\mathscr E'\_K(\mathbb R^n)$ iff
$\hat u$ can be extended to an entire function on $\mathbb C^n$ such that
$$
\vert\hat u(\zeta)\vert\le C\_0(1+\vert \zeta\vert)^{N\_0} e^{2π I\_K(\Im \zeta)},
$$
where $I\_K$ is the so-called supporting function of $K$. So
$\mathscr E'\_K(\mathbb R^n)$ appears as a subspace of $\mathscr E'(\mathbb R^n)$
which is characterized by the behaviour of its Fourier transform and not as a dual space.


