 Solution

The Neumann series criterion states that $\Vert A\Vert <1$ for a bounded linear operator on a Banach space implieswith convergence in the operator norm. In particular, $\Vert T\Vert <1$ suffices for invertibility of $I-T$.

If $|\lambda |>\Vert T\Vert$, apply this criterion to $A=T/\lambda$. It givesThus no such $\lambda$ belongs to the spectrum, and

For each integer $j\ge 1$, if $|\lambda {|}^j>\Vert T^j\Vert$, then ${\lambda }^{j}I-T^j$ is invertible by the same argument. The commuting factorizationthen gives a bounded two-sided inverse for $\lambda I-T$, namely the displayed sum multiplied by $({\lambda }^{j}I-T^j{)}^{-1}$. Hence every $\lambda \in \sigma (T)$ satisfies $|\lambda {|}^j\le \Vert T^j\Vert$. Taking the supremum over the spectrum and then the infimum over $j$ proves