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One question came to my mind while looking at the proof of Gelfand-Naimark theorem. Is Fourier transform a kind of Gelfand transform? Are there any other well-known transforms which are so?

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any.. i need some real examples of gefland transform –  K.Ghosh Dec 31 '12 at 11:17
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A good example is $L^1(\Bbb R)$ where the product is convolution. We need to add an unity element, hence we work on $L^1\times\Bbb R$ with the operation $(f_1,r_1)+(f_2,r_2)=(f_1+f_2,r_1+r_2)$ and $(f_1,r_1)*(f_2,r_2)=(f_1*f_2+r_1f_2+r_2f_1,r_1r_2)$. We can give a characterization of the homomorphisms: there are $$h_t(f,r):=\int_{\Bbb R}e^{its}f(s)ds$$ and $h_{\infty}(f,r)=r$. –  Davide Giraudo Dec 31 '12 at 11:20
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up vote 6 down vote accepted

Yes, the Fourier transform is the Gelfand transform for the commutative Banach algebras $(L_1(G),*)$ where $G$ is a locally compact commutative group. One can show that there is a bijection between the homomorphisms from $(L_1(G),*)$ into $\mathbb{C}$ and the elements of the Pontryagin dual $\hat{G}$, which is used to define the Fourier transform.

Some more details are in Kaniuth's book "A Course in Commutative Banach Algebras" (section 2.7).

See also the wikipedia entry on Pontryagin duality.

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