C star spectral radius equals norm for normal elements

For every normal $d\in A$ one has $\mathrm{\parallel }{d}^2\mathrm{\parallel }=\mathrm{\parallel }d{\mathrm{\parallel }}^2$: by [L1] and [L2], $\mathrm{\parallel }{d}^2{\mathrm{\parallel }}^2=\mathrm{\parallel }(d^2{)}^{\ast }(d^2)\mathrm{\parallel }=\mathrm{\parallel }{d}^{\ast }{d}^{\ast }dd\mathrm{\parallel }$ and, since $d^{\ast }d=d{d}^{\ast }$, the element $d^{\ast }{d}^{\ast }dd=(d^{\ast }d)(d{d}^{\ast })=(d^{\ast }d{)}^2$; so $\mathrm{\parallel }{d}^2{\mathrm{\parallel }}^2=\mathrm{\parallel }(d^{\ast }d{)}^2\mathrm{\parallel }=\mathrm{\parallel }(d^{\ast }d{)}^{\ast }(d^{\ast }d)\mathrm{\parallel }=\mathrm{\parallel }{d}^{\ast }d{\mathrm{\parallel }}^2=\mathrm{\parallel }d{\mathrm{\parallel }}^4$, whence $\mathrm{\parallel }{d}^2\mathrm{\parallel }=\mathrm{\parallel }d{\mathrm{\parallel }}^2$.

By induction on $n\ge 0$, $\mathrm{\parallel }{a}^{2^n}\mathrm{\parallel }=\mathrm{\parallel }a{\mathrm{\parallel }}^{2^n}$ for the normal element $a$: the case $n=0$ is $\mathrm{\parallel }a\mathrm{\parallel }=\mathrm{\parallel }a\mathrm{\parallel }$; if $\mathrm{\parallel }{a}^{2^n}\mathrm{\parallel }=\mathrm{\parallel }a{\mathrm{\parallel }}^{2^n}$, then $a^{2^n}$ is normal (a power of a normal element commutes with its adjoint, since $a$ and $a^{\ast }$ commute), so [step 1.1] applies to $d=a^{2^n}$ and gives $\mathrm{\parallel }{a}^{2^{n+1}}\mathrm{\parallel }=\mathrm{\parallel }(a^{2^n}{)}^2\mathrm{\parallel }=\mathrm{\parallel }{a}^{2^n}{\mathrm{\parallel }}^2=\mathrm{\parallel }a{\mathrm{\parallel }}^{2^{n+1}}$.

By [L3] the limit $r(a)={\lim }_k\mathrm{\parallel }{a}^k{\mathrm{\parallel }}^{1\mathrm{/}k}$ exists, and the sequence $k=2^n$ is a strictly increasing sequence of indices, so the subsequence $\mathrm{\parallel }{a}^{2^n}{\mathrm{\parallel }}^{1\mathrm{/}{2}^n}=\mathrm{\parallel }a\mathrm{\parallel }$ converges to $r(a)$; hence $r(a)=\mathrm{\parallel }a\mathrm{\parallel }$.