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I'm reading through Folland's Abstract Harmonic Analysis and I've come to a bit of a road block with some machinery developed for the Fourier inversion theorem. We know that the Fourier transform of a function in $L^1(G)$ need not be in $L^1(\widehat{G})$. In order to develop the Fourier inversion theorem, we need to restrict our domain to a proper subset of $L^1(G)$ so that the Fourier transform on this subset gives functions in $L^1(\widehat{G})$. Heuristically, this is very reasonable but the means by which we extract this subset is quite a mystery to me.

Given $\mu\in M(\widehat{G})$ (i.e. a finite complex Radon measure on $\widehat{G}$), we can define, for $g\in G$, the following

$$\phi_{\mu}(g) = \int_{\widehat{G}} \chi(g)\,d\mu(\chi).$$

This is a bounded continuous function on $G$. From here, we can define the set

$$\mathcal{B}(G) = \{\phi_{\mu}:\mu\in M(\widehat{G})\}.$$

Folland claims that the proper setting for the Fourier inversion theorem is on $\mathcal{B}(G) \cap L^1(G)$. I can't seem to justify to myself why we would even consider $\mathcal{B}(G)$ beyond the fact that $\phi_{\mu}$ somewhat looks like an inverse Fourier transform of $\mu$. What is the intuitive reasoning behind introducing $\mathcal{B}(G)$?

The Fourier inversion theorem on $\mathbb{R}$ is often done by considering bounded approximate identities a la the Gaussian or Cauchy functions but I can't see how this relates in any way $\mathcal{B}(G)$. As far as I am concerned at this point, $\mathcal{B}(G)$ is a black box that somehow facilitates the proof but has no real fundamental justification other than that it gets the job done.

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  • $\begingroup$ Your integral should be over $\widehat{G}$, not $G$. $\endgroup$ – KCd Jun 25 '14 at 3:40
  • $\begingroup$ Thanks for spotting the typo! $\endgroup$ – Cameron Williams Jun 25 '14 at 3:42
  • $\begingroup$ $\phi_{\mu}$ is the Fourier transform of $\mu$, not just "somewhat like". If $\mu$ is absolutely continuous w.r.t the Haar measure on $\hat{G}$, then the density of $\mu$ lies in $L^1(\hat{G})$. If you go back to the case of $\mathbb{R}$, this becomes a special case: Fourier transform can be inverted if one restricts to its preimage of $L^1$. $\endgroup$ – Michael Jun 25 '14 at 7:32

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