When does the spectrum of an element in a Banach algebra with involution lie in the open right half-plane?

Question

Let $A$ be a Banach algebra with involution, $x\in A$ and $t\in {\mathbb R}$ such that $t>\rho(xx^*)$. Show that $\sigma(te-xx^*)$ lies in the open right half-plane.

I have no idea! It's obvious that $t-\rho(xx^*)>0$. Maybe we should use this to conclude $\rho(te-xx^*)>0$ and then getting the statement! Would you please help me with that?

Answer 1

I think you have the right idea. We can replace $xx^*$ with $y$, as the anatomy of $xx^*$ is irrelevant.

Note that, \begin{align*} &s \in \sigma(y) \\ \iff \, &(y - se)^{-1} \text{ doesn't exist} \\ \iff \, &(se - y)^{-1} \text{ doesn't exist} \\ \iff \, &(te - y - (t - s)e)^{-1} \text{ doesn't exist} \\ \iff \, &t - s \in \sigma(te - y). \end{align*} If $s \in \sigma(y)$, then $$\Re(t - s) = t - \Re(s) \ge t - |s| > 0.$$