The enveloping $C^*$ algebra of $C^k[0,1]$. We know that $C^k[0,1]$ is an abelian $*$-Banach algebra, but it is not a $C^*$ algebra in general unless $k=0$.
I wonder what's the enveloping $C^*$ algebra of $C^k[0,1]$? I guess it might be $C[0,1]$. However, I am not able to prove it right now.
What I can show right now is that the maximal ideal space of $C^k[0,1]$ is $[0,1]$, In another word, all the $1$ dimensional $*$-representation are in this form.
In order to find the enveloping $C^*$ algebra, I think we need to find all the $*$-representation instead of those in only $1$ dimension. However, I have no idea how to do this.
Any help will be truly grateful!
 A: Observe that $f\in C^k=:A$ is invertible if and only if $f$ does not vanish on $[0,1].$ Hence $\sigma_A(f)=f([0,1]).$
For any $*$-representation $\pi: A\to B(\mathcal{H})$ and $f\in A,$ the operator $\pi(f)$  is normal. Therefore $$\|\pi(f)\|=\max\{|z|\in \mathbb{C}\,:\,z\in \sigma(\pi(f))\}$$
In other words the norm is equal to the spectral radius. As $\pi$ is a homomorphism ( $\pi(1)=I$) we have $$\sigma(\pi(f))\subset \sigma_A(f)=f([0,1])$$
Therefore
$$\|\pi(f)\|\le \max_{0\le x\le 1}|f(x)|=\|f\|_\infty$$
For every $t\in [0,1]$ we have the one-dimensional representation $\pi_t(f)=f(t).$ Therefore
$$\sup_{0\le t\le 1}\|\pi_t(f)\|=\|f\|_\infty$$
Summarizing the norm of $f$ in the $C^*$-enveloping algebra of $C^k[0,1]$ is equal $\|f\|_\infty.$ Therefore this algebra is $*$-isometrically isomorphic to $C[0,1].$
Remark Another way is to use the fact that the norm on the enveloping algebra can be determined by irreducible representations of $A,$ i.e. one-dimensional in the commutative case.
