$a,b\in A$ are selfadjoint elements of $C^*$-algebras, such that $a\le b$, why is $\|a\|\le \|b\|$ Let $A$ be a unital $C^\ast$-algebra with unit $1_A$.
a) Why is $a\le \|a\|1_A$ for all selfadjoint $a\in A$ and
b) If $a,b\in A$ are selfadjoint such that $a\le b$, why is $\|a\|\le \|b\|$?
I would try it with the continuous functional calculus $$\phi:C(\sigma(a))\to A,\; f\mapsto f(a),$$ for a) with $f(x)=|x|-x$, which is a positive function on $\sigma(a)\subseteq \mathbb{R}$, is it correct?
for b) It is $a\le \|a\|$ and $b\le \|b\|$, therefore $a\le \|b\|$. How can I continue?
Regards

For a) you don't need functional calculus. Note that $\|a\|\,1_A-a$ is a selfadjoint with spectrum contained in $[0,\infty)$, so positive ($\sigma(a+k\,1_A)=\{\lambda+k:\ \lambda\in\sigma(a)$).
As stated, b) is not true: $-3<-2$, but $\|-2\|=2<\|-3\|$. It is true for positives, though. For $0\leq a \leq b$, we represent $A\subset B(H)$, then 
$$
\langle a\xi,\xi\rangle\leq\langle b\xi,\xi\rangle
$$
for all $\xi\in H$. Then
$$
\|a\|=\sup\{|\langle a\xi,\xi\rangle:\ \|\xi\|=1\}
\leq
\sup\{|\langle b\xi,\xi\rangle:\ \|\xi\|=1\}
=\|b\|.
$$
If you want to do this without representing, the same idea can be achieved using states. 
