I have sincerely tried this problem, for way too long, and I must admit defeat. How am I to prove the following?
Let X be a Banach space and I be the identity mapping on X. If T is a bounded linear operator from X to X and $||I-T||<1$, then T is invertible and $\sum_0^\infty (I-T)^n$ converges to $T^{-1}$.
I have seen some proofs, but every single one uses $||(I-T)^n|| \le ||I-T||^n$ which is not true in general, or is it?
It is true that $\| A \circ B \| \le \| A \| \| B \| $ if you take supremum norm. Just consider
$$ X \overset B \to Y \overset A \to Z $$
all of the above being normed spaces. Then you have $$ \| A \circ B \| = \sup_{\| x \| \le 1} \| A B x \| \le \sup_{\| x \|\le 1} \|A \| \| B x \| = \| A \| \sup_{\| x \| \le 1} \| B x \| = \| A \| \| B \| $$ now apply induction to show $\| A^n \| \le \| A \|^n $.
I hope those proofs are complete now.
Regards,
D