Neumann Series Theorem

Theorem
Let $X$ be a Banach space.

Let $\map {CL} X$ be the continous linear transformation space.

Let $\norm {\, \cdot \,}$ be the supremum operator norm.

Let $A \in \map {CL} X$ be such that $\norm A < 1$.

Let $\circ$ be the composition of mappings.

Let $I$ be the identity mapping.

Then:


 * $I - A$ is invertible in $\map {CL} X$


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


 * $\norm {\paren{I - A}^{-1} } \le \dfrac 1 {1 - \norm A}$

Proof
Let $\ds S_k := \sum_{n \mathop = 0}^k A^n$.

$\ds \sum_{n \mathop = 0}^\infty A^n$ converges absolutely
By Supremum Operator Norm on Continuous Linear Transformation Space is Submultiplicative:


 * $\forall n \in \N : \norm {A^n} \le \norm A^n$

By assumption, $\norm A < 1$.

By Sum of Infinite Geometric Sequence, $\ds \sum_{n \mathop = 0}^\infty \norm A^n$ converges.

By series comparison, $\ds \sum_{n \mathop = 0}^\infty \norm {A^n}$ converges too.

By definition, $\ds \sum_{n \mathop = 0}^\infty A^n$ is absolutely convergent.

$\ds \sum_{n \mathop = 0}^\infty A^n$ converges
By assumption, $X$ is Banach.

By Necessary and Sufficient Conditions for Continuous Linear Transformation Space to be Banach Space, $\map {CL} X$ is Banach too.

Let $\ds S := \sum_{n \mathop = 0}^\infty A^n$.

We have that Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach.

Then $S$ converges in $\map {CL} X$.

In other words:


 * $\ds \lim_{k \mathop \to \infty} S_k = S$

Inverse of $\paren {I - A}$ is $S$
We have that:

Furthermore:

Hence:

That is:


 * $S A = A S = S - I$

It follows that:

Therefore, $I - A$ is invertible in $\map {CL} X$.

Furthermore:


 * $\ds \paren {I - A}^{-1} = S = \sum_{k \mathop = 0}^\infty A^k$

Moreover:

Also see

 * Definition:Neumann Series