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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!

Yanyu
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1 Answers1

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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.

Ryszard Szwarc
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  • The norm in $C^k$ is different from the norm in $C$, it should be defined here. https://math.stackexchange.com/questions/2291235/cn0-1-is-not-a-c-algebra – Yanyu May 09 '22 at 08:41
  • If not so, we assume that$C^k$ equipped with the norm $||\cdot||_{\infty}$, then it may not be a Banach space. Despite that, I think this proof still works. But how do we prove the remark you pointed out at last? I think this proof you give here is strongly related to the structure of continuous functions. Would you like to give me any reference for the general property you pointed out? – Yanyu May 09 '22 at 08:49
  • To be more specific, I think the equation $\sigma_A(f)=f([0,1])$ is trivial here, but we can't get a similar result if we get rid of the structure of continuous functions. – Yanyu May 09 '22 at 08:55
  • If the norm is different then my answer useless and I will delete it. Concerning your last question if $f\in C^k$ and $f$ does not vanish then $f$ is invertible in $C^k.$ So the spectra in $C[0,1]$ and $C^k[0,1]$ coincide. – Ryszard Szwarc May 09 '22 at 08:59
  • Although the norm is different, I think your proof still works for this special case. The key point is to realize that the spectra of $f$ in $C^k[0,1]$ is $f([0,1])$. I am not sure if your remark is true in general. – Yanyu May 09 '22 at 09:03
  • I have just re-edited the answer. This time I do not use the norm whatsover. – Ryszard Szwarc May 09 '22 at 09:07
  • I have removed the remark, as it became irrelevant. – Ryszard Szwarc May 09 '22 at 09:12