My question is whether Dirac-type distributions over an Abelian group define a basis of the Schwartz-Bruhat space $\mathcal{S}(G)^\times$ of tempered distributions on $G$, so that any distribution $f\in\mathcal{S}(G)^\times$ can be expressed as an integral of Dirac deltas
$$f=\int_X \mathrm{d}\mu(x)\,\, f(x)\, \delta_x $$
for some subset $X$ of $G$, some measure $\mu$ on $X$ and some function $f:X\rightarrow \mathbb{C}$.
I have attempted to solve this myself without much success. (I am a quantum physicist and lack background in harmonic analysis.) You can take a look at my attempt and a at more precise version of my question below.
Let $G$ be a Hausdorff locally compact Abelian group, $\mathcal{H}=L^2(G)$ the Hilbert space of two-integrable functions with its usual inner product. Let $\mathcal{S}(G)$ be the Schwartz-Bruhat space [ 2, 3 ] of smooth functions of rapid decay over $G$ and $\mathcal{S}(G)^\times$ its continuous dual vector space of tempered distributions, which I want to view it as a $\mathbb{C}$-vector space. The triple
$$\mathcal{S}(G)\subset \mathcal{H}\subset \mathcal{S}(G)^\times$$
is a rigged Hilbert space.
Question 1. One of the properties of the dual $\mathcal{S}(G)^\times$ is that it contains all Dirac-type delta distributions $\delta_g$ defined as
$$\langle \delta_g , \varphi\rangle = \varphi(g).$$
I would like to understand for which groups the set $\{\delta_g,\, g\in G\}$ defines a basis in $\mathcal{S}(G)^\times$. I assume already that $\mathcal{H}$ has a basis. The easy example is when $G$ is a finite Abelian group: in this case the $\delta_g$ represent Kronecker deltas and it can be shown that they define a basis of the dual space $\mathcal{S}(G)^\times$ using linear algebra.
Question 2. (A particular case.) If the answer to the above questions is negative I would be very interested to know what happens for groups of the form $$G=\mathbb{R}^a\times\mathbb{Z}^b\times \mathbb{R/Z}^c\times F$$ where $F$ is finite Abelian. This is actually the class of groups I am working with on a project. These groups are compactly-generated LCA groups, and are sometimes easier to work with.
My attempt. Since every tempered distribution $T$ can be obtained as the limit of a sequence $\{f_n\}_n$ of Schwartz-Bruhat functions, my (possibly wrong) intuition that I $\{\delta_g\}$ should define a basis of $\mathcal{S}(G)^\times$, because one should be able to write every function of the sequence as a linear combination of deltas exclusively
$$ f_n = \int_{G} \mathrm{d} g \,\, f_n(g)\,\,\delta_g$$
(where we integrate over the Haar measure of $G$), I think this should imply that $f$ is a linear combination of deltas in the limit, due to the fact that in the limit
$$\lim_{n\rightarrow \infty}\langle f_n , \varphi \rangle= \langle f, \varphi \rangle$$
for every test function $\varphi \in \mathcal{S}(G)$. But honestly, I do not know whether this that this argument is 100% correct and neither I know whether one can formalize it. I wonder also whether this type of argument could work for the class of groups $G=\mathbb{R}^a\times\mathbb{Z}^b\times \mathbb{R/Z}^c\times F$.
Best Answer
The short answer is that the $\delta_x$'s alone are not enough for a "basis". But it is enough to simply include all possible derivatives $\partial\delta_x$, $\partial\partial \delta_x$, ..., as well.