spectrum of some operators

Question

Suppose $A$ is an positive invertible operator in $B(H)$,$\{A_n\}$ is a sequence of positive invertible operators which converges to $A$ in norm topology. Can we conclude that $\log(\sigma(A_nA^{-1})) \to \{0\}$?

Answer 1

The "log" has no effect for the question you've asked. I can only assume your source included it because of some other purpose. Since the logarithm is continuous, $\operatorname{diam} \log(\sigma(A_nA^{-1})) \to 0$ if and only if $\operatorname{diam} \sigma(A_nA^{-1}) \to 0$.

Since $A_n \to A$, we have $A_nA^{-1} \to I$. By upper semi-continuity of the spectral diameter, $$\limsup_n \operatorname{diam} \sigma(A_nA^{-1}) \le \operatorname{diam} \sigma(I) = \operatorname{diam} \, \{1\} = 0$$ But since the diameter is $\ge 0$ by definition, it follows that $\operatorname{diam} \sigma(A_nA^{-1}) \to 0$.