Example 4

Consider the complex Banach algebra A of absolutely convergent Fourier series (example 8 of I, p. 19). For every element $u$ of the unit circle $\mathbf{U}$, the mapping $f\mapsto f(u)$ is a character ev$_u$ of A. If $f_0\in A$ is the identity mapping of $\mathbf{U}$, we have ev$_u(f_0) =u$, hence the mapping ev$:u\mapsto$ ev$_u$ of $\mathbf{U}$ into $\mathsf{X}(A)$ is injective; it is continuous.

Let $\chi \in \mathsf{X}(A)$. We have $\|f_0\|$ = $\|f_0^{-1}\|$ = 1, hence $|\chi (f_0)|\leqslant$ 1 and $|\chi (f_0)^{-1}|\leqslant 1$. This shows that $\chi (f_0)\in \mathbf{U}$ and there exists $u\in \mathbf{U}$ such that $\chi (f_0) =$ ev$_u(f_0)$. Since $\{f_0, f_0^{-1}\}$ generates the unital algebra A topologically, we have $\chi =$ ev$_u$. Thus, the mapping ev is a homeomorphism of $\mathbf{U}$ onto $\mathsf{X}(A)$, by which these spaces are identified. The Gelfand transform of A is then identified with the inclusion of A in $\mathscr{C}(\mathbf{U})$.

Since A is isomorphic to the Banach algebra $L^1(\mathbf{Z})$ (example 8 of I, p. 19), the space $\mathsf{X}(L^1(\mathbf{Z}))$ is identified with $\mathbf{U}$ and, for every element $(c_n)\in L^1(\mathbf{Z})$, the Gelfand transform $\mathscr{G}_{L^1(\mathbf{Z})}((c_n))$ is identified with the function $u\mapsto \sum_{n\in\mathbf{Z}}c_nu^n$ on $\mathbf{U}$.