Why does $\|(op)\|_{L^2{\infty} \rightarrow L^2{\infty}} \leq \frac{1}{2}$ $\implies$ $op$ is invertible?

Question

Why does

$\|(I+qG_{\zeta})\|_{L^2{\infty} \rightarrow L^2{\infty}} \leq \frac{1}{2}$

imply that $I+qG_{\zeta}$ is invertible operator?

Answer 1

Perhaps I'm confused by your notation, but as stated it is false. What if $q G_\zeta = -I$?

Perhaps what you mean is, if $\|T\| < 1$ (as a linear operator from any Banach space to itself), then $I+T$ is invertible. In fact, $(I+T)^{-1} = \sum_{n=0}^\infty (-T)^n$ which is a convergent series.