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Let $A$ be a unital $C^*$-algebra and $a\in A$ such that $r(a) < 1$. Define b = $(\sum_{n=0}^\infty (a^*)^n a^n)^{1/2}$. We can prove that $b\geq e$ and that $b$ is invertible. I want to show $\| b a b^{-1} \| < 1$.

From the definition of $b$ we see that $a^* b^2 a = b^2-e$ and we know $r(bab^{-1}) = r(a) <1$.

So it suffices to prove $r(b a b^{-1}) = \| b a b^{-1} \|$. It can follow from the fact $c = ba b^{-1}$ is a normal element... I don't know how to prove it (I have tried to compare $c^* c$ and $c c^*$...).


Context: The question appears in Murphy's book, page 74. I have managed to prove the first part. The second part of the question is to prove $$r(a)= \inf_{c\in Inv(A)}\{\|cac^{-1} \| \}$$ It's easy to see $$r(a) \leq \inf_{c\in Inv(A)}\{\|cac^{-1} \| \}$$ But I can't prove $$r(a) \geq \inf_{c\in Inv(A)}\{\|cac^{-1} \| \}$$ If we have had $r(a) = \|b a b^{-1} \|$ then it was obvious... But this is not true, so how we can prove this inequality?

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    (Jonas is right, btw.)2012-05-04
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    @JonasMeyer: I think this question is one of the exercises in Murphy's book on $C^\ast$-algebras2012-05-04
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    If this question has been taken from Murphy's book, then this should be acknowledged in the question. Likewise if this forms part of an exercise set or other homework.2012-05-04
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    $r(bab^{-1})$ is not typically $\|bab^{-1}\|$ (e.g. take $a$ nilpotent); $c=bab^{-1}$ is not typically normal. But $c^*c$ is a good thing to consider: it simplifies considerably, and using $b\geq e$ allows you to show that $\|c^*c\|<1$.2012-05-04
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    @YemonChoi:Thanks.2012-05-04
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    Jonas, Thanks for the hint. If we know only $\|b a b^{-1} \| < 1 $ and not $\|b a b^{-1} \| < r(a)$ I don't understand how to solve the second part of the question: $r(a)= \inf_{c\in Inv(A)}\{\|cac^{-1} \| \}$. how can I prove the inequality $r(a) \geq \inf_{c\in Inv(A)}\{\|cac^{-1} \| \}$ ?2012-05-04
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    Yes, The question appears in Murphy's book - page 74. link: http://imageshack.us/photo/my-images/152/murphy.png/2012-05-04
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    colge: Thanks for providing the source. Some notes on your first comment: (1) You didn't ask about showing anything about infs. If you wanted to ask about that, it should be included in your question. (2) You do not want to show that $\|bab^{-1}\|; that would in fact contradict what the second part of the problem asks you to show.2012-05-05
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    Jonas, I have just edited the question.2012-05-05
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    colge: It seems $r(a)=0$ should be treated as a special case. If $r(a)>0$, consider applying the first part to $\frac{1-\varepsilon}{r(a)}a$.2012-05-05

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