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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?

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$\|(I - T)^n\| \le \|I - T\|^n$ is definitely true in general. (Proof: you have $\|A^nx\| \le \|A\|\|A^{n-1}x\| \le ... \le \|A\|^n\|x\|$. Now take $A = I-T$.) – user38355 Apr 3 '13 at 21:18
you might consider reading up a little about Banach algebras while you are at it... – MBM Apr 3 '13 at 21:21
This is the first time I have dealt with normed spaces, so hank you. Given the definition, why is your first inequality true? That is, why is $||A^2x|| \le ||A||*||Ax||$? – Just Some Old Man Apr 3 '13 at 21:30
Ahhh, because Ax is a scalar. I get it now. – Just Some Old Man Apr 3 '13 at 21:38
It is true in essence because the operator norm is a Banach algebra norm: i.e. $\|ST\|\leq\|S\|\|T\|$. Every induced operator norm has this property. That's a good exercise. – 1015 Apr 3 '13 at 21:45
up vote 2 down vote accepted

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.



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