Does the inequality $0\leq a\leq b$ in a C*-algebra imply $\|a\|\leq\|b\|$? In relation to this question of mine: C* algebra inequalities
I am wondering if it is true that if $0\leq a \leq b$ in a C* algebra, does one have $||a||\leq||b||$?  If you need the C* algebra to be unital you can assume that since I could use the C* algebra unitization.  If it is not true, please state any reasonable conditions under which it is true because I see things like this used all the time, and it is possible that it merely applies in those special cases.
 A: Yes it is true in general.  As you mention, the question isn't affected by going to the unitization, so assume that the algebra is unital. 
Note that $b\leq \|b\|1$, because $\sigma(b)\subseteq[0,\|b\|]$, so that $\sigma(\|b\|1-b)\subseteq[0,\|b\|]$.  Thus $a\leq \|b\|1$, and since $\|a\|$ is in the spectrum of $a$, this implies that $\|a\|\leq \|b\|$.

You could use Gelfand theory here, because $a$ and $b$ both commute with $\|b\|1$, so you could consider $C^*(b,1)$ then $C^*(a,1)$.  However, all that is used above are the facts that $x\geq 0$ if and only if $x=x^*$ and $\sigma(x)\subseteq [0,\infty)$, and if $x\geq 0$, then the spectral radius of $x$ is $\|x\|$ (true for all normal elements, and quickly derived from the spectral radius formula and the C*-identity).  Regardless, the "trick" is to use $\|b\|1$ as an intermediary between the possibly noncommuting $a$ and $b$.
A: When $a$ and $b$ commute, it is true. You only need to look at the commutative C*-subalgebra generated by $a, b, 1$. But for the general case, I don't remember any thing right now.  
