# Spectral radii and norms of similar elements in a C*-algebra: $\|bab^{-1}\|<1$ if $b=(\sum_{n=0}^\infty (a^*)^n a^n)^{1/2}$

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?

• (Jonas is right, btw.) – user16299 May 4 '12 at 1:47
• @JonasMeyer: I think this question is one of the exercises in Murphy's book on $C^\ast$-algebras – user16299 May 4 '12 at 2:37
• 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. – user16299 May 4 '12 at 5:08
• $r(bab^{-1})$ 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 $\|c^*c\|<1$. – Jonas Meyer May 4 '12 at 10:49
• @YemonChoi:Thanks. – Jonas Meyer May 4 '12 at 10:51

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 $\|x+y\|\geq \|x\|-\|y\|$.