The other day I proved that if $f \in \ell^1 (\mathbb Z)$ then its Gelfand transform $\widehat{f}$ is a map $S^1 \to S^1$ such that $$ \widehat{f}(z) = \sum_{k \in \mathbb Z}f(k) z^k$$
and that $f(k)$ are the Fourier coefficients of $\widehat{f}$. So the statement I proved is that $\widehat{f}$ equals its Fourier series. By coincidence, this $\widehat{f}$ is also the Fourier transform of $f$: the Fourier transform of an $f: \mathbb R\to \mathbb C$ is defined to be $$ \widehat{f}(e^{-2 \pi i \xi}) = \int_{-\infty}^\infty f(x) e^{-2 \pi i x \xi}dx$$
but since in the case of $\ell^1$ the domain is $\mathbb Z$ the integral becomes a sum and $$ \widehat{f}(e^{-2 \pi i k }) = \sum_{k \in \mathbb Z} f(k) e^{-2 \pi i k } $$
That is, the Fourier transform of $f \in \ell^1 (\mathbb Z)$ equals the Fourier series of $\widehat{f}$. This is not true in an arbitrary setting. Hence my question: when does the Fourier transform $\widehat{f}$ of $f$ equal the fourier series of $\widehat{f}$?