$a,b\in A$ 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

let A be a C*-algebra and a,x in A with 0≤xx*≤a.
questions:
1-  Put bn=(((1/n)+a)^(-1/2))(a^(1/4))x. show that (bn) is a cauchy sequence in A.
2-  Show that the limit b in A satisfies x=(a^(1/4))b and ∥b∥≤∥a∥^(1/4)
for question 1- it is easy but for question 2- how i show that ∥b∥≤∥a∥^(1/4)?
