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

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How do you define the integral of distribution valued functions on $G$? –  Qfwfq Jan 17 at 14:27
    
@Qfwfq: There are several possibilities for vector-valued integrals, for example the Pettis-Integral. –  Johannes Hahn Jan 17 at 15:04

1 Answer 1

up vote 2 down vote accepted

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.

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3  
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". –  Igor Khavkine Jan 17 at 14:19
    
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? –  Juan Bermejo Vega Jan 18 at 10:31
1  
A reference: W.Rudin, 6.26. –  Sergei Akbarov Jan 18 at 11:32
    
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 $0$. –  Jochen Wengenroth Jan 18 at 12:15
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@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. –  Igor Khavkine Jan 18 at 13:30

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