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Let's say that in a unital C* algebra, we have $b \geq a \geq 0$ and $a$ is invertible. Then $b$ is also invertible. Can we conclude that $a^{-1} \geq b^{-1}$? If so, why? Can any related statement be made if we only assume $a \leq b$? (And maybe that $a$, and hence $b$ are self adjoint.)

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This is an abstract version of this earlier question:… (Essentially the same answer applies. And it was applied by the same answerer.) – Jonas Meyer Dec 15 '12 at 5:40
up vote 3 down vote accepted

Yes. If $b \ge a \ge 0$ and $a$ is invertible, then $a^{-1/2} b a^{-1/2} \ge a^{-1/2} a a^{-1/2} = 1$ so $ a^{1/2} b^{-1} a^{1/2} = (a^{-1/2} b a^{-1/2})^{-1} \le 1$ and then $b^{-1} = a^{-1/2} a^{1/2} b^{-1} a^{1/2} a^{-1/2} \le a^{-1}$. Without the $\ge 0$ it's false, e.g. try $a = -1$ and $b = 1$.

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actually I saw that argument but I just don't understand why the special case you use in there holds. You're saying that if $a \leq 1$ and $a$ is invertible, then $a^{-1} \geq 1$. (and vice versa) – Jeff Aug 17 '12 at 7:35
Oh wait now I finally see. The way to see this fact is to use the functional calculus, and compare $t$, $1/t$, and the constant function $1$ defined on the spectrum which is a compact subset of $(0, 1]$. – Jeff Aug 17 '12 at 7:38
@Jeff: You can also show that directly, just using the spectrum, if you defined $b ≥ a$ by $\mathrm{spec}(b-a) \subseteq [0, ∞)$, which I believe is how it is commonly done for C*-algebras. – canaaerus Jul 3 '15 at 8:14

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