Why does the order on positive elements respect the order on the norm?

Question

The question title doesn't quite convey what I mean, but close enough. I'm struggling with a bit of Davidson's "C* algebras by example", in his proof of Lemma 1.4.7. Our hypotheses: $0 \leq A \leq B$ are invertible elements of a $C^*$ algebra, where $X \geq Y$ means that $X-Y$ is a positive element. We encounter the following sentence:

Thus $(A^{1/2}B^{-1/2})^*(A^{1/2}B^{-1/2}) \leq I$; whence $\|A^{1/2}B^{-1/2}\| \leq 1$. The adjoint has the same norm, so $$I \geq (A^{1/2}B^{-1/2})(A^{1/2}B^{-1/2})^* = A^{1/2}B^{-1}A^{1/2}$$

I'm lost on this. Why does the "whence" actually follow?

Answer 1

The assertion says that $X^*X\leq I$ implies $\|X\|\leq 1$. First note that $\|X\|^2=\|X^*X\|$, so it is enough to show "if $0\leq X\leq I$, then $\|X\|\leq 1$".

Next you note that $0\leq X\leq I$ implies that $\sigma(X)\subset[0,1]$. And finally, using that $X$ is selfadjoint, $\|X^2\|=\|X\|^2$, which you use to show that $\|X\|$ equals the spectral radius of $X$. Then $\|X\|\leq1$.

Answer 2

Let $f(t) = t^2$, then $f(t)$ is a non-negative operator monotone function on $[0, \infty)$.

There is a theorem that says $f(||X||) \leq ||f(|X|)||$ when $f(t)$ is a non-negative

operator monotone function and $||.||$ is a normalized unitarily invariant norm.

Now, the statement at the beginning of $\star$ means that $|A^{1/2}B^{-1/2}|^2 \leq I$,

hence, $f(||A^{1/2}B^{-1/2})||) \leq ||f |A^{1/2}B^{-1/2}| || \leq ||f(I)|| = 1$

which implies that $||A^{1/2}B^{-1/2}|| \leq 1$.