Invertibility of Identity Minus Operator

Theorem
Let $\GF \in \set {\R, \C}$.

Let $X$ be a Banach space over $\GF$.

Let $T : X \to X$ be a bounded linear operator such that:


 * $\norm T_{\map \BB X} < 1$

where $\norm {\, \cdot \,}_{\map \BB X}$ denotes the norm of a bounded linear operator.

Then $I - T$ is invertible as a bounded linear operator.

In particular:


 * $\ds \paren {I - T}^{-1} = \sum_{n \mathop = 0}^\infty T^n$

Proof
For each $n \in \N$, define:


 * $\ds S_n = \sum_{k \mathop = 0}^n T^k$

We argue first that $\sequence {S_n}_{n \mathop \in \N}$ is convergent.

Since $X$ is a Banach space, it is enough to show that $\sequence {S_n}_{n \mathop \in \N}$ is Cauchy.

Let $\epsilon > 0$.

Let $m, n \in \N$ with $n > m$.

Then we have:

From Sum of Infinite Geometric Sequence, we have:


 * $\ds \sum_{k \mathop = 0}^\infty \norm T_{\map \BB X}^k$

since $\norm T_{\map \BB X} < 1$.

So:


 * $\ds \sequence {\sum_{k \mathop = 0}^n \norm T_{\map \BB X}^k}_{n \mathop \in \N}$ is Cauchy.

So we can pick $N \in \N$ such that for $n > m \ge N$ we have:


 * $\ds \sum_{k \mathop = m + 1}^n \norm T_{\map \BB X}^k < \epsilon$

So for $n > m \ge N$ we have:


 * $\norm {S_n - S_m}_{\map \BB X} < \epsilon$

Hence from we have:


 * $\norm {S_n - S_m}_{\map \BB X} < \epsilon$

for all $n, m \in \N$ with $n, m \ge N$.

So $\sequence {S_n}_{n \mathop \in \N}$ is Cauchy and hence convergent in $\map \BB X$.

Denote:


 * $\ds S = \lim_{n \mathop \to \infty} S_n = \sum_{k \mathop = 0}^\infty T^k$

To finish we show that $\paren {I - T} S = S \paren {I - T} = I$.

Now for each $n \in \N$, we have:

and similarly:

We now show that $I - T^{k + 1} \to I$ in $\map \BB X$.

We have:

On the other hand by Convergence of Product in Normed Algebra, we have:


 * $S_n \paren {I - T} \to S \paren {I - T}$

and:


 * $\paren {I - T} S_n \to \paren {I - T} S$

in $\map \BB X$, while:


 * $S_n \paren {I - T} = \paren {I - T} S_n = I - T^{k + 1} \to I$

So we have:


 * $S \paren {I - T} = \paren {I - T} S = I$