$a,b\in A$ 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

Prologue
The spectrum is not linear in general for several reasons.
So the trick is to reduce the inequality involving only the operator and the identity.
Solution
Spectral relations:
$$\sigma(A+1)=\sigma(1)+\sigma(A)\quad\sigma(\lambda A)=\lambda\sigma(A)$$
By Neumann series:
$$N^*N=NN^*:\quad\|\sigma(N)\|=\|N\|$$
By spectral relations:
$$A=A^*:\quad-\|A\|1\leq A\leq\|A\|1$$
Both together give:
$$0\leq A\leq B\implies 0\leq A\leq\|B\|1\implies0\leq\sigma(A)\leq\|B\|\implies\|A\|\leq\|B\|$$
Concluding the proof.
Epilogue
No need of continuous calculus - it is completely elementary!
The proof is based on the very definition of the ordering merely.
