Approximate Identities

Question

Statement : Let $U$ be a C* algebra and$\lambda=\left\{A\in U:A\geq 0, ||A||<1\right\}$. If $B\in \lambda$ then if $X\in U$

$$||X^*(I-B)^2X||\leq ||X^*(I-B)X||$$

For reference this is from Davidson's $C^*$ Algebras by Example, where he proves that every $C^*$ algebra has an approximate identity.

Question how do they arrive at that inequality?

Answer 1

Check that: $$0\leq(I-B)^2\leq(I-B)$$ (Not sure about that point!)

Then apply: $$0\leq A\leq B\implies X^*AX\leq X^*BX$$

Exploit that: $$0\leq A\leq B:\quad\|A\|\leq\|B\|$$ (That is crucial of positive elements!)

Answer 2

Using the ordering suggested by Freez_S we get that $$ (I-B)^2\leq (I-B)$$This is the same as showing $$0\leq B(I-B)$$ after a little bit of rearrangement.

Now we see this in terms of the continuous functional calculus. We note that in the sub algebra $C^*(B)$ there exists an element associated to the continuous function $f(x)=x(1-x)$ namely $f(B)=B(I-B)$. Since the spectra of this element is contained in the image of $f(x)$ on the interval $[0,1]$ we see that $B(I-B)$ has non-negative spectra. Since the spectra of $B$ in our original algebra $U$ is a subset of the spectra with respect to $C^*(B)$; it follows that $B(I-B)$ is a positive element, the self adjointness being obvious.