about the Beurling Theorem in Murphy's book

Question

Here is a part of Murphy's book C*-Algebras and Operator Theory:

In the fourth line of his proof, he claims that $1-\lambda a$ is invertible. But we do not have $||\lambda a|| <1$, how to show invertibility of $1-\lambda a$? Thanks!

Answer 1

You have $\lambda^{-1} -a$ invertible because $|\lambda^{-1}|>r(a)$ (since $|\lambda|<1/r(a)$). Then $$ 1-\lambda a=\lambda\,(\lambda^{-1}-a) $$ is invertible. The case $\lambda=0$ is trivial.