I was told to check that in this example the Gelfand representation is non-isometry nor surjective. So I'm trying to write a proof for it.
My attempt:
Our Gelfand representation is $\phi: A \rightarrow C_0(\Omega(A))$, defined by $a \rightarrow \hat{a}$, where $\Omega(A)$ is the set of characters on $A$.
Show it's non-isometry: we want to show $\|\phi a\|= \| \hat{a}\| \ne \| a\|$. This is equivalent to showing that the map is not injective, i.e $Ker(\phi)\ne \{0 \}$. I feel like somehow I need to use that if $f' \in A$, then $f' \in C([0,1])$ to show the kernel is not zero. But I'm not sure what to do from here.
Show it's not surjective: We want to find $\hat{a} \in C_0(\Omega(A)) $ such that nothing in $A$ will be mapped to $\hat{a}$. I'm not sure how to proceed, Again I feel like I'm missing something from that $f' \in C([0,1])$.
Any help will be appreciated.
It is not hard to see that the arguments that show that $\Omega(C[0,1])$ are the pointwise evaluations still work for $\Omega(A)$.
Now you can take the $x$ from the example (the identity). You have $\|x\|=2$, $\|\hat x\|=1$. So the Gelfand transform is not an isometry. It is injective, though.
If $f\in C[0,1]$ is not differentiable, it cannot be the image of some $g\in A$. This is because the characters are the point evaluations: if $f=\phi(g)$ we would have, for any $a\in[0,1]$, $$ g(a)=a(g)=\hat g(a)=f(a). $$ So $\phi$ is not surjective.