Positive Elements and Inequalities in a $C^*$-algebra

Question

The following corollary comes from Fillmore's book A User's Guide to Operator Algebras

Corollary 2.3.5 $0 \le a \le b \implies ||a|| \le ||b||$

Here is the proof:

We have $b \le ||b||$ by 2.3.1, and so $a \le ||b||$ by the theorem. Hence $r(a) \le ||b||$ and therefore $||a|| \le ||b||$ by 2.1.3

Why is it the case that $r(a) \le ||b||$? I was unable to verify this.

Answer 1

Since $0\leq \|b\|-a$, we have $\sigma(\|b\|-a)\subset[0,\infty)$ (here $\sigma(x)$ denotes the spectrum of $x$) and thus $\sigma(a)\subset(-\infty,\|b\|]$. But $a$ is positive, so $\sigma(a)\subset[0,\infty)$, and therefore $\sigma(a)\subset[0,\|b\|]$, i.e., $r(a)\leq\|b\|$.