I have a question concerning exercise 6 in chapter 2 of Murphy's $C^*$-algebras and Operator Theory. The exercise is as follows:
Let A be a unital $ C^* $-algebra. If $r(a) < 1$ and $b = (\sum_{n=0}^{\infty}(a^*)^n a^n)^{\frac{1}{2}}$, show that $b \geq 1$ and $\|bab^{-1}\| < 1$.
So, I have been able to prove that $b$ is greater than 1, but I am having trouble with the second part. A hint was given here: Spectral radii and norms of similar elements in a C*-algebra: $\|bab^{-1}\|<1$ if $b=(\sum_{n=0}^\infty (a^*)^n a^n)^{1/2}$
I tried simplifying $(bab^{-1})^*bab^{-1}$ as suggested, but have not been able to make any progress. I can see that $a^*b^2a$ simplifies to $b^2$, but I cannot make the connection to the rest. Any more help would be greatly appreciated.
Edit: Correction to the last paragraph, $a^*b^2a$ simplifies to $b^2-1$.
As you noted, $a^*b^2a=b^2-1$. Then $$ \|bab^{-1}\|^2=\|(bab^{-1})^*bab^{-1}\|=\|b^{-1}a^*b^2ab^{-1}\| =\|b^{-1}(b^2-1)b^{-1}\|=\|1-b^{-2}\|. $$ The element $b^{-2}$ is positive, and since $b\geq1$, we get that $b^{-2}\leq 1$. Then $0\leq 1-b^{-2}\leq1$, which implies that $\|1-b^{-2}\|\leq 1$. But we can say a bit more. Since $b\geq1$, we know that $\sigma(b)\subset [1,\beta]$ for some $\beta\geq1$. Then $\sigma(1-b^{-2})\subset \big[0,1-\frac1{\beta^2}\big]$. Thus $$ \|bab^{-1}\|^2=\|1-b^{-2}\|=1-\frac1{\beta^2}<1. $$