Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a Hausdorff locally-compact Abelian group and $L^2(G)$ the Hilbert space of two-integrable complex functions on the group.

Question. What would be natural vector space $\mathcal{R}$ of complex functions such that its dual vector space $\mathcal{R}^\times$ of continuous complex linear functionals fulfill the following properties?:

(1) $\mathcal{R}^\times$ contains all Dirac delta measures of the group, i.e. the functionals $$\langle \delta_g , \varphi\rangle = \varphi(g),\quad \textrm{ for all }\, g\in G,\,\varphi\in\mathcal{R}.$$

(2) Any functional in $f\in\mathcal{R}^\times$ can be expressed as some integral of Dirac deltas $$f=\int_X \textrm{d}\mu(x)\,\, f(x)\, \delta_x $$ for some subset $X$ of $G$, some (suitable) measure $\mu$ on $X$ and some (suitable) complex function $f$.

(3) In (2) there is flexibility to choose the allowed classes of measures $\mu$ and complex functions $f$ as long as $$\mathcal{R}\subset L^2(G)\subset \mathcal{R}^\times$$ is a rigged Hilbert space.

Motivation

A space $\mathcal{R}$ chosen as above would be a *minimal rigged Hilbert space*$\mathcal{R}\subset L^2(G)\subset \mathcal{R}^\times$ that is wide enough to contain the Dirac measures as distributions but tight in the sense that any other object in $\mathcal{R}^\times$ can be written as a "linear combination" (an integral as in property 1.) of Dirac deltas. In this case, one can think of the set $\{\delta_g\}$ as an (infinite) basis of the vector space $\mathcal{R}^\times$. My original motivation was to understand for what kind of distribution spaces (over groups) the Dirac deltas $\{\delta_g\}$ define such a "basis". (See also this previous post of mine.)

Can $\mathcal{R}$ have nice dual properties?

Since $G$ is an LCA group we can consider its dual group of continuous character functions $\widehat{G}$; then Plancherel's theorem tells us that $L^2(G)$ is isomoprhic to $L^2(\widehat{G})$ via the Fourier transform of $G$. For applications in harmonic analysis, it would be desirable that the properties (1-3) of $\mathcal{R}^\times$ defined above were preserved under the application of the Fourier transform. In particular, I wonder whether $\mathcal{R}$ could have a 4th additional property. (Perhaps it follows easily from the previous ones?)

(4) $\mathcal{R}^\times$ contains all Dirac delta measures of the dual group $\widehat{G}$ (the group continuous character functions of $G$). In other words, the functionals $$\langle \delta_{\chi} , \varphi\rangle = \langle \chi, \varphi \rangle = \widehat{\varphi}(\chi),\quad \textrm{ for all }\, \chi\in \widehat{G},\,\varphi\in\mathcal{R}.$$

Here, $\mathcal{\varphi}$ denotes the Fourier transform of $\varphi$, which is a function in $L^2(\widehat{G})$

share|improve this question
    
When you talk of continuous linear functionals on $\mathscr{R}$ you tacitly assume that $\mathscr{R}$ is a topological vector space. What topology are you thinking of? (The space of continuous compactly supported functions will do the trick.) –  Liviu Nicolaescu Jan 20 at 14:35
    
I am just interested on constructing the rigging, so, I would take the simplest topology such that the inclusions in the containment $\mathrm{R}⊂L^2(G)⊂\mathrm{R}^\times ×$ are continuous. Would that be the subspace topology in $L^2(G)$? –  Juan Bermejo Vega Jan 23 at 7:52
    
The space you mention is what I called $C_c(G)$. I know that there is a version of the [Riesz–Markov–Kakutani] theorem showing that $C_c(G)^\times$ is precisely the space of regular Borel measures on $G$. However, I do not know how to apply this result: are such measures "integral combinations" of Dirac measures? –  Juan Bermejo Vega Jan 23 at 8:00
    
At the very beginning I was actually hoping that merely taking the space of continuous (or, perhaps, uniformly continuous) functions in $L^2(G)$ would be enough. (Let's call such spaces $C(G)$ and $C_u(G)$.) Given that, I was hoping that taking $\mathcal{R}$ to be $C(G)\cap C(\widehat{G})$ or $C_u(G)\cap C_u(\widehat{G})$ would give you a dual $\mathcal{R}^\times$ containing all Dirac measures of $G$ and $\widehat{G}$ and "only them". I could not work this out, though; in particular, I do not if $\mathcal{R}$ would be dense. –  Juan Bermejo Vega Jan 23 at 8:09
1  
There is a result valid for any locally compact group: the convolution of any Borel regular measure with the Dirac measure at the origin is the original measure. If you unravel the definition of convolution of measure you deduce that any measure is a "superposition" of $\delta$'s or, as you put it, an "integral" combination of $\delta$'s –  Liviu Nicolaescu Jan 23 at 13:39
add comment

1 Answer 1

Let me suggest that the duality that you might be after is that developed by Jan Pachl between the space of bounded, uniformly continuous functions on $G$ and the uniform measures thereon. In order to get the appropriate dual space, one uses on the former not the norm topology (the dual would then be too large) but the structure introduced by Pachl. The latter has recently presented a comprehensive version of this theory (which goes back more than thirty years) in the monograph "Uniform spaces and measures". Those aspects of his theory which are related to harmonic analysis were described in arXiv 0608139. (In the first part of your question, the group structure seems only to be relevant in order to specify the measure--- presumably, Haar measure).

The space of uniform measures on $G$ contains the Dirac measures in a very special way. Their linear hull is dense. In fact, the former is the free complete locally convex space over $G$ in the following sense (from category theory): every uniformly continuous, bounded mapping from the former into a Banach space has a unique extension to a continuous linear mapping. This might be the property that you are looking for.

I don't see how you can expect the kind of symmetry you are hoping for in the second part (consider the special case of the duality beteen the circles group and the integers) and the condition about having a rigged Hilbert space would only be valid in the weaker form that you would replace $L^2$ by the space of locally square integrable functions.

I hope this is of some relevance to your query (which was a tad vague, after all).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.