Proposition 6.4.3.label If $A$ is a $C^{*}$-algebra, then $\norm{x}{}=\norm{x}{sp}$ for every normal $x\in A$.
Proof. We go straight to computation
\begin{align*}\norm{x^{2n}}{}^{2}=\norm{(x^{2n})^*x^{2n}}{}=\norm{(x^*x)^{2n}}{}=\norm{(x^*x)^n(x^*x)^n}{}=\norm{(x^*x)^n}{}^{2}\end{align*}
applying this iteratively
\begin{align*}\norm{x^{2^n}}{}=\norm{(x^*x)^{2^{n-1}}}{}=\norm{(x^*x)^{2^{n-2}}}{}^{2}=\cds=\norm{x^*x}{}^{2^{n-1}}=\norm{x}{}^{2n}\end{align*}
By Proposition 6.2.6
\begin{align*}\norm{x}{}=\limit{n\to\infty}\norm{x^{2^n}}{}^{2^{-n}}=\norm{x}{sp}\end{align*}
$\square$