This answer states the Fourier transform is the Gelfand transform on the Banach algebra $L^1(G)$ with convolution. I've read the resource linked in the answer, but I still have some confusion.

My first questions is simply how can this be true? Wouldn't the Gelfand transform of $f\in L^1(G)$ (let's denote it $\Gamma(f)$) simply operate as $\Gamma(f)(\chi) = \chi(f)$? This doesn't seem to be how the Fourier transform acts, which instead maps $\chi\mapsto(f*\chi)(1)$. I don't see clearly how these mappings relate to one another.

Perhaps this is explained by the bijection $\widehat{G}\to\Delta(L^1(G))$ mapping $\chi\mapsto(f\mapsto\hat{f}(\chi))$ (which I believe I do understand) but I don't see how to make this work either.

My second question is somewhat contingent on the answer to the first, but I'll try my best. Letting $A(\widehat{G}) = \{\hat{f}:f\in L^1(G)\}$, if I accept that the Fourier transform is somehow the Gelfand transform, I would immediately know that $A(\widehat{G})$ is dense in $C(\widehat{G})$. Since I don't yet believe that Fourier=Gelfand, can I easily deduce this from Gelfand-Naimark?

  • $\begingroup$ I would not call the Gelfand transform ”Fourier transform”, but on the reals they are the same. $\endgroup$ – AD. May 6 at 12:48

For your first question: Try to determine which are the bounded $\ast$-homomorphisms $\chi:L^1(G) \to \mathbb C$ for $G$ Abelian. Observe that if $G$ is discrete taking $g \mapsto \chi(\delta_g)$ give you a character (and this is a bijection). If $G$ is not discrete you would have to use "approximations of identity" or look up Chapter 3/Section 3.2 of

Folland, Gerald B., A course in abstract harmonic analysis, Studies in Advanced Mathematics. Boca Raton, FL: CRC Press. viii, 276 p. (1995). ZBL0857.43001.

For your second question: Use Stone-Weierstrass theorem. You'll have to show that $A(G)$ separates points, which can be done using that if $f, g \in L^2(G)$ have supports $E$, $F$, then $f\ast g$ has support contained in $E F$.

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