$b \le \|b\|$ even when $b$ is not normal or self-adjoint?

Question

It is a theorem in $C^\ast$ algebras that if $0\le a \le b$ then $\|a\|\le \|b\|$. The proof given in this book (page 47) starts by asserting that $b \le \|b\|$ because we can use the Gelfand transform on the $C^\ast$ algebra generated by $1$ and $b$.

But the Gelfand theorem only applies to commutative algberas and here $b$ is neither assumed to be normal nor self-adjoint so the algebra generated by $1$ and $b$ is not commutative.

Hence my questions:

(1) Is there an assumption missing in the statement of the theorem?

(2) Or is it possible to prove that $0 \le a \le b$ implies $0\le b$?

Answer 1

What you need to know is that a positive operator $a$ can be written as $a = s^*s$ for some element $s$ in the $C^*$-algebra. Particularly, it means that every positive element is indeed self-adjoint since $a^* = (s^*s)^* = s^*s = a$ so that $a^*a = aa^*$. This in turn gives you that the $C^*$-algebra of a positive element is commutative since $a$ commutes with $a^*$. (This follows since the $C^*$ algebra generated by $a$ is basically the set of polynomials in $a$ and $a^*$.) Particularly, since $0\le a\le b$, $0\le b$ and so $b$ is positive, thus you can employ Gelfand as desired to get that $b\le \|b\|$.