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I am reading a proof of this theorem:

If $a,b$ are positive elements of a $C^\ast$ algebra and $a \le b$ then $a^{1/2}\le b^{1/2}$.

I don't understand one step in the proof. I understand this: Let $t > 0$ and $c,d$ be such that $c + i d = (t + b + a)(t + b - a)$. Then $c \ge t^2$ therefore $c$ is positive and invertible.

Given this, why is $1 + i c^{-1/2}dc^{-1/2}$ invertible?

  • I have a feeling that you can show that $|c^{-1/2}dc^{-1/2}|<1$ which would ensure that what you have is invertible. – Cameron Williams Dec 01 '14 at 04:28
  • It was my first thought but I didn't see how to do it. –  Dec 01 '14 at 04:29
  • Well we know that $|c^{-1}|\le t^{-2}$. If you can show that $|d| < t^2$, you'd be good to go. – Cameron Williams Dec 01 '14 at 04:39
  • @CameronWilliams Yes I know. I have of course thought of that, too. But again I didn't see how to do it. So if you know how to prove it please post an answer, I will appreciate it. –  Dec 01 '14 at 05:35

1 Answers1

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Here is the argument:

Since $d$ is selfadjoint, so is $c^{-1/2}dc^{-1/2}$. Then the spectrum of $1+ic^{-1/2}dc^{-1/2}$ is of the form $$ \{1+i\lambda:\ \lambda\in\sigma(c^{-1/2}dc^{-1/2})\}. $$ The $\lambda$ are real, so the spectrum does not contain $0$, and thus $1+ic^{-1/2}dc^{-1/2}$ is invertible.

As this equals $c^{-1/2}(c+id)c^{-1/2}$ we conclude that $c+id$ is invertible. Now one can conclude that $t+b-a$ is invertible. As this happens for all $t>0$, $b-a\geq0$, i.e. $a\leq b$.

Martin Argerami
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  • This is putting the cart before the horse. You use the fact in OP's statement to prove that $t+b+a$ and $t+b-a$ are invertible... – Cameron Williams Dec 01 '14 at 05:29
  • What do you mean? I haven't seen the rest of the proof the OP mentions, but the assumption seems to be $0\leq a\leq b$. – Martin Argerami Dec 01 '14 at 05:30
  • In the proof, you use the fact that $c+id$ is invertible to show that $t+b+a$ is invertible. It can be found in Murphy. – Cameron Williams Dec 01 '14 at 05:33
  • Dear Martin, thank you very much. It really seems easy after reading your answer! The book is really weird, it argues the other way around: it states because the LHS of your second displaystyle equation is invertible it follows that the RHS is also invertible and since $c$ is invertible therefore $c + id$. But I think that's purposefully obscurely written and your argument is much better. –  Dec 01 '14 at 05:33
  • @student: Cameron is right. The proof in Murphy tries to show that $a^2\leq b^2$ implies $a\leq b$. So my argument does not apply to that proof. Please Unaccept so that I can delete the answer. – Martin Argerami Dec 01 '14 at 05:36
  • But you showed $c + id$ is invertible. That's all the proof needs. What am I missing? –  Dec 01 '14 at 05:38
  • Everything. The proof does not assume that $a\leq b$. It assumes that $a^2\leq b^2$. Then you cannot reason as I did. – Martin Argerami Dec 01 '14 at 05:39
  • Bleh, you're right. I kept looking at the statement of the theorem but really the first line of the proof starts with $a^2 \le b^2 $. How annoying. Sorry for the confusion. –  Dec 01 '14 at 05:40
  • I think I got it. Please see the edit. – Martin Argerami Dec 01 '14 at 05:42
  • Yep! Perfect. I feel so silly. I can't believe I didn't notice that the spectrum basically got bumped off of the real axis. – Cameron Williams Dec 01 '14 at 05:47