Show that (c) $S=T^{-1}.$, (d)$\Vert S_{n}-T^{-1}\Vert\leq \frac{\lambda^{n+1}}{1-\lambda}.$, (e) $S_{n+1}=I+(I-T)S_{n}.$

Question

For the following question, I need just hint not a whole solutions. As you see I have solved the first two parts.

The Problem: Let $X$ be a Banach space and $T : X \to X$ be linear with $\Vert I −T\Vert = \lambda < 1.$ Let $S_n=\sum\limits_{j=0}^{n}(I-T)^{j}.$ Show that

(a) T is one to one. (I have solved)

(b) The sequence $\{S_{n}\}_{n=0}^{\infty}$ is convergent in $\mathcal{B}(X).$ (I have solved)

(c) Let $S=\lim\limits_{n\to \infty}S_{n}.$ Show that $S=T^{-1}.$

(d) $\Vert S_{n}-T^{-1}\Vert\leq \frac{\lambda^{n+1}}{1-\lambda}.$

(e) $S_{n+1}=I+(I-T)S_{n}.$

Answer 1

(c) Show that $\lim\limits_{n\to \infty}S_{n}T=\lim\limits_{n\to \infty}TS_{n}=I$.

(d) Note that $S-S_n=(I-T)^{n+1}S$

(e) This is obvious.

Answer 2

I'd do this by doing (e) first, then (d), then observing the right hand side of (d) $\to 0$.

(e) should be clear by definition. Then, by induction, we can show (d). Hints: (base case, use the reverse triangle inequality, inductive step use (e)).

I feel like this is the most natural way of doing these questions, since (c) is an easy consequence of (d) which is a straightforward consequence of (e).