# How exactly is the Fourier transform the same as the Gelfand transform?

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?

• I would not call the Gelfand transform ”Fourier transform”, but on the reals they are the same. – 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
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 $$E F$$.