Invertibility of Identity Minus Operator

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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:

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


Corollary

Let $T : X \to X$ be a invertible bounded linear operator.

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

$\norm S_{\map \BB X} \norm {T^{-1} }_{\map \BB X} < 1$


Then $T + S : X \to X$ is an invertible bounded linear operator.


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:

\(\ds \norm {S_n - S_m}_{\map \BB X}\) \(=\) \(\ds \norm {\sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^m T^k}_{\map \BB X}\)
\(\ds \) \(=\) \(\ds \norm {\sum_{k \mathop = m + 1}^n T^k}_{\map \BB X}\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = m + 1}^n \norm T_{\map \BB X}^k\) Norm Axiom $\text N 2$: Positive Homogeneity, Norm on Bounded Linear Transformation is Submultiplicative: Corollary

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 Norm Axiom $\text N 2$: Positive Homogeneity 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:

\(\ds \paren {I - T} S_n\) \(=\) \(\ds \paren {I - T} \paren {\sum_{k \mathop = 0}^n T^k}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^n T^{k + 1}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 1}^{n + 1} T^k\)
\(\ds \) \(=\) \(\ds I - T^{n + 1}\)

and similarly:

\(\ds S_n \paren {I - T}\) \(=\) \(\ds \paren {\sum_{k \mathop = 0}^n T^k} \paren {I - T}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^n T^{k + 1}\)
\(\ds \) \(=\) \(\ds I - T^{k + 1}\)

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

We have:

\(\ds \norm {\paren {I - T^{k + 1} } - I}_{\map \BB X}\) \(=\) \(\ds \norm {T^{k + 1} }_{\map \BB X}\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(\le\) \(\ds \norm T_{\map \BB X}^{k + 1}\) Norm on Bounded Linear Transformation is Submultiplicative: Corollary
\(\ds \) \(\to\) \(\ds 0\) since $\norm T_{\map \BB X} < 1$

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:

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

$\blacksquare$

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