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

Question

Can I use this problem:

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 $(I - T)^{-1} = \sum_{n=1}^\infty{}T^n.$

To prove this one

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.

If so, How?

Kavi Rama Murthy · Accepted Answer · 2020-03-19T08:48:50.003

1

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

edited Mar 19 '20 at 08:48

answered Mar 19 '20 at 08:40

Kavi Rama Murthy

359,332

So we need not use the question I mentioned? – Intuition Mar 19 '20 at 08:44
Why the first part of your first sentence leads you to say that "so $T^{-1}S$ is invertible" I can not see this .... could you clarify please? – Intuition Mar 19 '20 at 08:47
but why convergence means invertible? as 0 operator is convergent but it is not invertible. – Intuition Mar 19 '20 at 09:01
@Secretly You are allowed to use the fact that $|U|<1$ implies $I-U$ is invertible (and its inverse is given by a certain series). Take $U=I-T^{-1}S$. – Kavi Rama Murthy Mar 19 '20 at 09:07

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

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