normal elements in spectral theory.

If $x$ is a normal element in a $C^{*}-algebra$ then $x$ is unitary $\iff$ $\sigma(x) \subseteq (the \ unit \ circle \ in \ the \ complex \ plane)$. According to the book I am reading this should be "obvious" according to continuous functional calculus. I somehow understand the proof of this direction $\implies$. However, is it possible to show the other direction without using functional calculus, the direction where we assume $\sigma(x) \subseteq (the \ unit \ circle \ in \ the \ complex \ plane)$ and then show it forces $x$ to be unitary?

Solution 1:

Let me start by saying that I see absolutely no reason of avoiding the functional calculus. Trust me: it is your best friend in a $C^*$-algebra. However, here is an argument.

I will use that the spectrum in a unital abelian $C^*$-algebra described as

$$\sigma(x)=\{\phi(x):\phi:A\to\mathbb{C}\text{ is a nonzero *-hom}\}$$

and that, if $B\subset A$ is a unital inclusion between $C^*$-algebras, then $\sigma_B(b)=\sigma_A(b)$ for all $b\in B$ (something more general is true, but this suffices for our purposes). You can find these facts in any elementary operator algebras textbook, such as Murphy.

Let $A$ be a unital $C^*$-algebra and let $a\in A$ be a normal element with $\sigma(a)\subset\mathbb{T}$.

Now let $z\in\sigma(a^*a)$. Since $a$ is normal, $C^*(a,1_A)$ is a unital abelian $C^*$-algebra, unitally included in $A$. Also, $a^*a\in C^*(a,1_A)$. By the fact that $\sigma_A(a^*a)=\sigma_{C^*(a,1_A)}(a^*a)$ and the description of the spectrum in a unital, abelian $C^*$-algebra, namely in $C^*(a,1_A)$, we find a $*$-homomorphism $\phi:C^*(a,1_A)\to\mathbb{C}$ so that $\phi(a^*a)=z$. But then $z=\phi(a^*)\phi(a)=\overline{\phi(a)}\phi(a)=|\phi(a)|^2$. But again by our description and the invariance of spectrum in unital subalgebras we have $\phi(a)\in\sigma_{C^*(a,1_A)}(a)=\sigma_A(a)\subset\mathbb{T}$, so $|\phi(a)|^2=1$ and thus $z=1$. We have thus shown that $\sigma_A(a^*a)\subset\{1\}$ and we know that the spectrum is non-empty in Banach algebras, so $\sigma_A(a^*a)=\{1\}$. But this means $a^*a=1_A$, showing that $a$ is a unitary.