Character Space of $C^1[0,1]$ and Gelfand Representation I have recently been working through some exercises in Murphy's "C*-Algebras and Operator Theory," and I am having some trouble with Exercise 10 in Chapter 1. The exercise is as follows:
Let $A = C^1[0,1]$. Let $x: [0,1] \longrightarrow \mathbb{C}$ be the inclusion. Show that $x$ generates $A$ as a Banach algebra. If $t \in [0,1]$, show that $\tau_t$ belongs to $\Omega(A)$, where $\tau_t$ is defined by $\tau_t(f) = f(t)$, and show that the map $[0,1] \longrightarrow \Omega(A)$, $t \mapsto \tau_t$, is a homeomorphism. Deduce that $r(f) = \|f\|_{\infty}$ ($f\in A$). Show that the Gelfand representation is not surjective for this example.
So far, I have been able to show that $x$ generates $A$ via Stone-Weierstrass. The claim that $\tau_t$ is in the character space seemed quite clear as well. I am having some trouble with the rest. To show the homeomorphism, it should be enough to show surjectivity, since continuity and injectivity should be clear, but I am not sure how to approach this. I am also not quite sure how to approach the last two claims either. Any help would be appreciated.
 A: The canonical way to do this is to use that given $\omega\in\Omega(A)$, its kernel $\ker\omega$ is a maximal ideal on $A$. Then you show that maximal ideals in $A$ are of the form $\ker\omega=\{f:\ f(t_\omega)=0\}$ for a fixed $t_\omega\in[0,1]$. From there you get that
$$
\omega(f)=\omega(f-f(t_\omega)1)+f(t_\omega)=f(t_\omega), 
$$
so $\omega=\tau_{t_\omega}$. In the context of Murphy's book, you can also use Theorem 1.3.7, that tells you that $\Omega(A)\simeq[0,1]$ via the map $\beta:\omega\longmapsto \omega(x)$. This map is the inverse of $\tau$, since
$$
\tau_{\omega(x)}(f)=f(\omega(x))=\omega(f).
$$
As for the rest, from the Gelfand Representation Theorem you have that
\begin{align}
r(f)&=\|\hat f\|_\infty=\sup\{\hat f(\omega):\ \omega\in\Omega(A)\}
=\sup\{\omega(f):\ \omega\in\Omega(A)\}\\[0.3cm]
&=\sup\{f(t_\omega):\ t_{\omega}\in[0,1]\}=\|f\|_\infty.
\end{align}
The Gelfand transform cannot be surjective because in this case it is the inclusion $C^1[0,1]\hookrightarrow C[0,1]$. Indeed, under the equivalent of $\Omega(A)$ and $[0,1]$,
$$
\hat f(t_\omega)=\omega(f)=f(t_\omega), 
$$
so $\hat f=f$. This last equality also gives a faster proof of $r(f)=\|f\|_\infty$.
