Theorem $:$
Let $(X,\|\cdot\|)$ be a Banach space. Let $T:X \longrightarrow X$ be a continuous linear transformation. Suppose $a=\|T\| < 1$. Then there exists $S:X \longrightarrow X$ such that $ST=TS=I$.
Proof $:$
Define for each $n \ge 0,$ $S_n = \sum\limits_{k=0}^{n} T^k = I + T +T^2 + \cdots + T^n$. Then $S_n$ is bounded and linear.
$$S_n (I-T) = \sum_{k=0}^{n} T^k (I-T) = {\sum_{k=0}^{n} T^k} - {T \sum_{k=0}^{n} T^k} = {\sum_{k=0}^{n} T^k} - {\sum_{k=0}^{n} T^{k+1}} = I - T^{n+1}$$
$\implies S_n (I - T) (x) = (I- T^{n+1}) (x)$.
$\implies x-S_n(I-T) (x) = T^{n+1} (x)$.
So,
$$\|x - S_n(I-T) (x) \| = \|T^{n+1} (x) \| \le \|T^{n+1}\| \|x\| \le \|T\|^{n+1} \|x\| = a^{n+1} \|x\| \rightarrow 0$$
$\implies S_n(I-T) (x) \rightarrow x$ as $n \rightarrow \infty$ for all $x \in X$. i.e. $S_n(I-T) \rightarrow I$ as $n \rightarrow \infty$. So, $(I-T) \sum\limits_{n=0}^{\infty} T^n = I$.
$\implies \sum\limits_{n=0}^{\infty} T^n = (I-T)^{-1}$.
This completes the proof.
I don't understand the proof. What does the result obtained in this proof signify? How do I find $S$ from here. Is $S = \sum\limits_{n=0}^{\infty} T^n$? If 'yes' then $ST=TS$ holds but it equals to $S-I$, not equals to $I$. Also I don't understand where the Banach space condition is used in the proof. Please help me in this regard.
Thank you very much.
Yes, $S=\lim_{n\to\infty}S_n=\sum_{n=0}^\infty T^n$. You need the fact that the space is complete in order to prove that this series converges. And, yes, $ST=TS\ne\operatorname{Id}$. My guess is that the statement of the theorem is that$$(\operatorname{Id}-T)S=S(\operatorname{Id}-T)=\operatorname{Id}$$because that's what was proved.