# C*-algebra norm comparable to sup-norm?

I am trying to show what amounts to a special case of the (commutative) Gelfand-Naimark theorem. That is: For a self-adjoint element in a unital C*-algebra A there exists a unique isomorphism $$C^*(1,a)\xrightarrow{\varphi}C(sp(a))$$

where sp(a) is the spectrum of $a$. If we let $\varphi(a)=x$ and $\varphi(1_A)=1$ then there is a unique extension of this to the algebra generated by $1$ and $a$. I run in to trouble showing that this extension is continuous. I think this is a question of whether the norm inherited from the C*-algebra is comparable to the sup-norm. The root of the problem seems to be that I am not able to conclude that the coefficients $\alpha$ and $\beta$ are necessarily small when $\|\alpha +\beta a\|$ is small.

Any suggestions would be appreciated.

• That the extension is continuous is because, for a self-adjoint element $a, \|a\| = r(a)$, the spectral radius, and the spectral radius is precisely the sup-norm on the right-hand-side. – Prahlad Vaidyanathan Nov 4 '15 at 5:31
• Yes but it is not immediately clear to me how a norm preserving operator on a generating set extends to a norm preserving or even a continuous operator on the whole algebra. Perhaps I am overlooking something fairly obvious though. – SchwarzWithNoT Nov 4 '15 at 7:17
• Well, it extends to the algebra of polynomials in $\{1,a\}$ in the obvious fashion, and it remains norm preserving there. And from there it extends to $C^{\ast}(1,a)$ by density (as one does for any liner map) – Prahlad Vaidyanathan Nov 4 '15 at 8:12

## 1 Answer

I'll be surprised if you can avoid the usual route of going through maximal ideals and characters.

Showing that your map is continuous is not a step in your problem but the whole problem. And what you want to show is that $\|p(a)\|=\|p|_{\sigma(a)}\|^{\vphantom \int}_\infty$ for any polynomial $p$. This requires understanding the structure of $\sigma(a)$, and the usual way is the equality $$\sigma(a)=\{\varphi(a):\ \varphi\text{ is a character }\}.$$