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.
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 $b$.
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.