Why convergence means invertibility?

Question

I was trying to understand the solution given in this question:

Using $(I - T)^{-1} = \sum_{n=1}^\infty{}T^n$ in the solution of another problem.

But I did not understand why convergence of the operator in the first question there means its invertibility? Especially because I know that the zero operator is convergent but it is not invertible. Could anyone explain this for me please?

Here is the first question:

Let $X$ be a Banach space and $T \in \mathcal{L}(X, X)$ with $\|T\| < 1.$ Define $T^0 = I.$ (a) Use the completeness of $\mathcal{L}(X, X)$ to show that $\sum_{n=1}^{\infty} T^n$ converges in $\mathcal{L}(X, X).$ (b) Show that $(I - T)^{-1} = \sum_{n=1}^\infty{}T^n.$

And here is the answer that was given to the second question based on the first one:

$\|I-T^{-1}S\|=\|T^{-1}T-T^{-1}S\| \leq \|T^{-1}\|\|T-S\|<1$ so $T^{-1}S$ is invertible by the result you have quoted. This implies that $S=T(T^{-1}S)$ is also invertible.

And here is the second question:

Let $X$ be a Banach space and $T,S \in \mathcal{L(X)}.$ Suppose that $T$ is invertible and that $\|T-S\| < \|T^{-1}\|^{-1}.$ Prove that $S$ is invertible.

score 2 · Accepted Answer · 2020-03-19T09:35:20.517

2

For any $n>0$, you have $(I-T)\sum\limits_{i=0}^nT^i = I - T^{n+1}$. As the norm of $T^n$ tends to $0$ when $n\to \infty$, it follows that $I-T$ is the inverse of the infinite sum.

edited Mar 19 '20 at 09:35

answered Mar 19 '20 at 09:25

I do not understand why I have "For any $n>0$, you have $(I-T)\sum\limits_{i=0}^nT^i = I - T^n$" could you explain more please? – Intuition Mar 19 '20 at 09:30
Also do you have a proof for this " As the norm of $T^n$ tends to $0$ when $n\to \infty$,"? – Intuition Mar 19 '20 at 09:32
The product is $I - T + T - T^2 + T^2 - \cdots - T^n +T^n - T^{n+1}$, but there was actually a typo. For every $0<k<n+1$, you have both $\pm T^k$, so all those cancel out. It should have given $I-T^{n+1}$, sorry about that. Fixed the typo. – Mar 19 '20 at 09:35
The norm thingy follows from convergence. If it didn't tend to 0, the norm of the sum would grow too rapidly. – Mar 19 '20 at 09:38
what was the typo? – Intuition Mar 19 '20 at 09:39
Do you have a proof for your last comment? – Intuition Mar 19 '20 at 09:39
$T^n$ was the typo. – Mar 19 '20 at 09:43
Are you there to answer my last 2 comments? – Intuition Mar 19 '20 at 09:43
Also, could you please look at that question if you have time https://math.stackexchange.com/questions/3585432/why-we-take-u-f-in-l2a-b-f-l2a-b-leq-1 – Intuition Mar 19 '20 at 09:44
https://en.wikipedia.org/wiki/Neumann_series read this. – Mar 19 '20 at 09:46

Why convergence means invertibility?

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