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 $G=\mathbb{R}^a\times\mathbb{Z}^b\times \mathbb{R/Z}^c\times F$.

  • $\begingroup$ How do you define the integral of distribution valued functions on $G$? $\endgroup$
    – Qfwfq
    Jan 17 '14 at 14:27
  • $\begingroup$ @Qfwfq: There are several possibilities for vector-valued integrals, for example the Pettis-Integral. $\endgroup$ Jan 17 '14 at 15:04
  • $\begingroup$ In the definition of distributions, there is that for any distribution $T \in D'(\mathbb{R})$ and for any interval $[a,b]$ there is a finite $k$ and a continuous function $f$ such that for any $\varphi \in C^\infty_c([a,b]), \langle T,\varphi \rangle = \langle f, \varphi^{(k)} \rangle$. Your post asks if always $k = 1$, the answer is no, it is only locally finite. $\endgroup$
    – reuns
    Sep 21 '18 at 18:13

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.

  • 3
    $\begingroup$ BTW, the term "basis" is not the correct precise term here. The concept you are using is analogous to a basis, but only if you replace "finite linear combinations" in the usual definition of a basis with "integrals with respect to certain measures". $\endgroup$ Jan 17 '14 at 14:19
  • $\begingroup$ I think I still do not understand why this happens, and why my limit argument seems to fail. Also, for practical purposes, could I just choose a new tighter rigging $\mathcal{R}_G$ fulfilling $\mathcal{S}_G\subset \mathcal{R}_G\subset \mathcal{H} \subset \mathcal{R}_G^\times \subset \mathcal{S}_G^\times$ and so that $\mathcal{R}_G^\times$ is precisely the space spanned by the integrals of Dirac deltas? $\endgroup$ Jan 18 '14 at 10:31
  • 1
    $\begingroup$ A reference: W.Rudin, 6.26. $\endgroup$ Jan 18 '14 at 11:32
  • $\begingroup$ A simple argument that not all tempered distributions can be represented as you want is that (for all reasonnable integrals) a distribution $f=\int \varphi(x) \delta_x d\mu(x)$ should be of order $0$ because all $\delta_x$ have order $0$. $\endgroup$ Jan 18 '14 at 12:15
  • 1
    $\begingroup$ @JuanBermejoVega, if you set up your $\mathcal{R}_G$ to consist of only continuous functions, then I think could get what you want. A convolution of a measure with $\delta_x$ is just the identity transformation. So you essentially want the dual space to $\mathcal{R}_G$ to consist of measures, and that's (one version) of the Riesz representation theorem. $\endgroup$ Jan 18 '14 at 13:30

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.