Is the range of the Gelfand transform closed?

Question

Let $A$ be a commutative unital Banach algebra. Consider the Gelfand map $\Gamma:A\longrightarrow C(M_A)$, $\Gamma(a)=\hat{a}$, where $M_A$ is the Character space of $A$. Is the image of the Gelfand map, $\Gamma(A)$ closed in $C(M_A)$ with the uniform norm?

Answer 1

Not in general: Take $A = C^1[0,1]$ of continuously differentiable functions on $[0,1]$ with the norm $\|f\| := \|f\|_{\infty} + \|f'\|_{\infty}$. For each $t\in [0,1]$, the evaluation map $\tau_t : A \to \mathbb{C}$ given by $f\mapsto f(t)$ induces a continuous map $$ [0,1] \to M_A \text{ given by } t\mapsto \tau_t $$ One can then show that this map is a homeomorphism (since $A$ is generated by the function $f_0(t) = t$) and so the Gelfand transform then is simply the inclusion map $$ A\to C(M_A) \cong C[0,1] $$ and so its image is dense but not closed in $C[0,1]$ with the uniform norm.