This is actually a continuation of my question here: Norm inequality of positive element of $C^*$-Algebra with norm less than 1, but i prefer to make it into new question.
So previously, i claimed that if $x, y$ are positive elements of unital $C^*$-Algebra with $||x+y||<1$, then $\dfrac{||x||}{1-||y||}<1$. Turns out it was not true in general. But then, i see this Lemma in some paper that you can access here: https://www.researchgate.net/publication/286009948_C-algebra-valued_metric_spaces_and_related_fixed_point_theorems.
It means, if $x, y \in \mathbb{A}^{'}_{+}$ with $||x+y||<1$, we have $$ \begin{equation} ||x+y||<1 \rightarrow x+y \preceq I \rightarrow x \preceq I-y \rightarrow x(I-y)^{-1} \preceq I \rightarrow ||x(I-y)^{-1}|| \leq 1 \end{equation} $$
I have two questions:
- Does $a$ have to commute so that lemma 2.1(iii) holds, i.e If $a \in \mathbb{A}$ (not necessarily commute element), is this inequality still holds: $(I-a)^{-1}c \preceq (I-a)^{-1}b$? I know that multiplication of two positive elements not always positive, maybe that is the case if we consider the invertible element, but i am not sure.
- From $||x+y||<1$, can we conclude that $||x(I-y)^{-1}|| < 1$ (strict inequality). I have tried this by contradiction. Suppose that it is possible to have $||x(I-y)^{-1}|| = 1$ when $||x+y||<1$, i hope that it will contradict either $||x||<1$ or $||y||<1$ but i can't get there. Is there any clue?
Thank you for your help.
Note: the paper said that the lemma is some results from Murphy and Douglas book, but i can't find it in those books. Maybe someone have a better eye than me, please let me know. Regards
Yes, the lemma requires that $a$ is in the centre of $\mathbb A$ (the notation $\mathbb A'$ is fairly unusual here, even if it kind of makes sense). In general the elements $(I-a)^{-1}b$ and $(I-a)^{-1}c$ will not be selfadjoint, and the order comparison makes no sense.
Yes, assuming that $I-y$ is invertible and that $y$ commutes with $x$. If $\|x+y\|<1$, then there exists $\delta>0$ with $\|x+y\|<1-\delta$. Then $$x+y\leq(1-\delta)I\implies x\leq (1-\delta)I-y=I-y-\delta I=(I-y)(I-\delta(I-y)^{-1}).$$ If $y$ commutes with $x$, $$\tag1 x(I-y)^{-1}\leq I-\delta(I-y)^{-1}. $$ Since $y$ is positive and $y\leq I-x$, we have that $0\leq y\leq I$, and to $0\leq I-y\leq I$. Then $(I-y)^{-1}\geq I$. This gives us $$\tag2 I-\delta(I-y)^{-1}\leq(1-\delta)I. $$ Combining $(1)$ and $(2)$, $$ \|x(I-y)^{-1}\|\leq 1-\delta<1. $$
As a general comment, the way to deal with inequalities in a C$^*$-algebra is by using that $$a\leq b\implies cac^*\leq cbc^*.$$ So when you have $x\leq I-y$ and both are positive with $I-y$ invertible, you get $$ (I-y)^{-1/2}x(I-y)^{-1/2}\leq I. $$