Theorem: Invertibility of Identity Minus Operator

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: 
{{begin-eqn}}
{{eqn	| l = \norm {S_n - S_m}_{\map \BB X}
		| r = \norm {\sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^m T^k}_{\map \BB X}
}}
{{eqn	| r = \norm {\sum_{k \mathop = m + 1}^n T^k}_{\map \BB X}
}}
{{eqn	| o = \le
		| r = \sum_{k \mathop = m + 1}^n \norm T_{\map \BB X}^k 
		| c = {{NormAxiomVector|2}}, Norm on Bounded Linear Transformation is Submultiplicative: Corollary
}}
{{end-eqn}}
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 {{NormAxiomVector|2}} 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: 
{{begin-eqn}}
{{eqn	| l = \paren {I - T} S_n
		| r = \paren {I - T} \paren {\sum_{k \mathop = 0}^n T^k}
}}
{{eqn	| r = \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^n T^{k + 1}
}}
{{eqn	| r = \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 1}^{n + 1} T^k
}}
{{eqn	| r = I - T^{n + 1}
}}
{{end-eqn}}
and similarly: 
{{begin-eqn}}
{{eqn	| l = S_n \paren {I - T}
		| r = \paren {\sum_{k \mathop = 0}^n T^k} \paren {I - T}
}}
{{eqn	| r = \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^n T^{k + 1}
}}
{{eqn	| r = I - T^{k + 1}
}}
{{end-eqn}}
We now show that $I - T^{k + 1} \to I$ in $\map \BB X$. 
We have: 
{{begin-eqn}}
{{eqn	| l = \norm {\paren {I - T^{k + 1} } - I}_{\map \BB X}
		| r = \norm {T^{k + 1} }_{\map \BB X}
		| c = {{NormAxiomVector|2}}
}}
{{eqn	| o = \le
		| r = \norm T_{\map \BB X}^{k + 1}
		| c = Norm on Bounded Linear Transformation is Submultiplicative: Corollary
}}
{{eqn	| o = \to
		| r = 0
		| c = since $\norm T_{\map \BB X} < 1$
}}
{{end-eqn}}
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: 
:$S \paren {I - T} = \paren {I - T} S = I$
{{qed}}
Category:Bounded Linear Operators
Category:Invertibility of Identity Minus Operator
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