Theorem 34.4.3.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then $\norm{x}_{A} = [x]_{sp}$.
Proof, [Theorem II.8.1, Zhu93]. First suppose that $x$ is self-adjoint. In this case,
\begin{align*}\normn{x^2}_{A}&= \normn{xx^*}_{A} = \norm{x}_{A}^{2} \\ \normn{x^{2^n}}_{A}&= \norm{x}_{A}^{2^n}\end{align*}
for all $n \in \natp$. Thus by the spectral radius formula,
\[[x]_{sp}= \limsup_{n \to \infty}\norm{x^{n}}_{A}^{1/n}\ge \limsup_{n \to \infty}\normn{x^{2^n}}_{A}^{1/2^n}= \norm{x}_{A}\]
Now suppose that $x$ is only normal. Since $x$ and $x^{*}$ commute, $[xx^{*}]_{sp}\le [x]_{sp}[x^{*}]_{sp}$ by Proposition 33.5.11. Thus
\[\norm{x}^{2}_{A} = \normn{xx^*}_{A} = [xx^{*}]_{sp}\le [x]_{sp}[x^{*}]_{sp}= [x]_{sp}^{2}\]
by (5) of Proposition 34.1.3.$\square$
Post a Comment