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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?

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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.

Prahlad Vaidyanathan
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  • Thank you. You are correct Prahlad. Because if it is closed, then $C^1[0,1]$ will become a Banach space with the uniform norm, which is not true. –  Jul 11 '16 at 06:53
  • sorry I'm noob. $f_0(t) = t$ generates the algebra $A$ because it generates the polynomials, right, so a character $\phi \in M_A$ is fully determined by $\phi(f_0)$. Then how do you know $\hat{f_0}$ generates $C(M_A)$ ? is it obvious here, or do I need first to prove the Gelfand representation in general ? – reuns Jul 11 '16 at 07:30
  • Yes, a character is determined by its value on $f_0$, so if $\varphi \in M_A$ and $\alpha = \varphi(f_0)$, you first show that $\alpha \in [0,1]$, so $\alpha = f_0(t_0)$ for some $t_0 \in [0,1]$. Hence, $\varphi$ and $\tau_{t_0}$ agree on a dense subset (of polynomials), and so must agree on $A$. This gives the homeomorphism between $M_A$ and $[0,1]$. – Prahlad Vaidyanathan Jul 11 '16 at 08:12