Gelfand transform

Remarks

Each $\hat a$ is continuous by the definition of the weak-* topology, and it vanishes at infinity. The map $\Gamma$ is an algebra homomorphism with $\norm{\hat a}_\infty \le \norm{a}$. If $A$ has a unit, then $\hat a(\Delta(A)) = \sigma(a)$, the spectrum of $a$, so $\norm{\hat a}_\infty$ is the spectral radius of $a$. The kernel of $\Gamma$ is the Jacobson radical of $A$.

Example: let $A = \ell^1(\Z)$ with convolution $(a * b)_n = \sum_k a_k b_{n-k}$. For $z$ in the unit circle, $\varphi_z(a) = \sum_n a_n z^n$ is linear, nonzero, and multiplicative by the Cauchy product formula, so it is a character. Every character has this form, and $z \mapsto \varphi_z$ identifies $\Delta(A)$ with the circle. The Gelfand transform of $a$ is the absolutely convergent Fourier series $\hat a(z) = \sum_n a_n z^n$.

Non-example: the construction needs $A$ to be commutative. The algebra $M_2(\C)$ has no characters at all. A character vanishes on commutators, and the commutators span the trace-zero matrices. So its kernel is a nonzero two-sided ideal of the simple algebra $M_2(\C)$, hence everything. So for $M_2(\C)$ the "Gelfand transform" would map into functions on the empty set and lose all information.