Intereting Posts

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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 $c c^*$…).

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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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The first part can be solved by simplifying $(bab^{-1})^*bab^{-1}$ to show that it has norm less than $1$, using the fact that $b\geq 1$. (In particular, $a^*b^2a$ simplifies nicely.)

The second part can be solved, in the case when $r(a)>0$, by applying the first part to $\frac{1-\varepsilon}{r(a)}a$, yielding for $0<\varepsilon<1$ an invertible $b$ such that $\|bab^{-1}\|\leq\frac{r(a)}{1-\varepsilon}$. The case when $r(a)=0$ can then be solved by applying the previous case to $a+\varepsilon e$ and using the inequality $\|x+y\|\geq \|x\|-\|y\|$.

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