Properties of Product of Identity plus Operator Raised to Powers of 2

Theorem
Let $X$ be a Banach space.

Let $\map \LL X$ be the set of all linear transformations.

Let $\map {CL} X$ be a continuous linear transformation sapce.

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

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

Let $I$ be the identity mapping.

Let $\circ$ be the composition of mappings.

For all $n \in \N$ let $A^n := \underbrace {A \circ A \circ \ldots \circ A}_{n \text{ times}}$, with $A^0 := I$.

For all $n \in \N$ let $P_n = \underbrace{\paren {I + A} \circ \paren {I + A^2} \circ \ldots \circ \paren {I + A^{2^n} } }_{n + 1 \text{ terms} }$

Then:


 * $\forall n \in \N : \paren {I - A} \circ P_n = I - A^{2^{n + 1} }$


 * the sequence $\sequence {P_n}_{n \mathop \in \N}$ converges in $\map {CL} X$ to $\paren {I - A}^{-1}$

$\paren {I - A} \circ P_n = I - A^{2^{n + 1} }$
This will be a proof by induction.

Basis for the induction
Let $n = 0$.

Then:

Induction hypothesis
The induction hypothesis is the following statement:


 * $\paren {I - A} \circ P_n = I - A^{2^{n + 1} }$

from which it is to be shown that:


 * $\paren {I - A} \circ P_{n + 1} = I - A^{2^{\paren {n + 1} + 1} }$

$\sequence {P_n}_{n \mathop \in \N}$ converges to $\paren {I - A}^{-1}$
We have that $\norm A < 1$.

Furthermore:

We have that Set of Linear Transformations with Supremum Operator Norm is Normed Vector Space.

Hence, $\sequence {I - A^{2^{n + 1} } }_{n \mathop \in \N}$ converges in $\struct {\map \LL X, \norm {\, \cdot \,}}$ to $I$.

By Neumann Series Theorem, $I - A$ is invertible in $\map {CL} X$.

Moreover:

Since:

Thus, $\sequence {P_n}_{n \mathop \in \N}$ converges to $\paren {I - A}^{-1}$.