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

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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 $\paren {I - A}^{-1}$


Proof

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

This will be a proof by induction.


Basis for the induction

Let $n = 0$.

Then:

\(\ds \paren {I - A} \circ P_0\) \(=\) \(\ds \paren {I - A} \circ \paren {I + A}\)
\(\ds \) \(=\) \(\ds I + A - A - A^2\)
\(\ds \) \(=\) \(\ds I - A^2\)
\(\ds \) \(=\) \(\ds I - A^{2^\paren{0 + 1} }\)


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:

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


Induction Step

\(\ds \paren {I - A} \circ P_{n + 1}\) \(=\) \(\ds \paren {I - A} \circ P_n \circ \paren {I + A^{2^{n + 1} } }\)
\(\ds \) \(=\) \(\ds \paren {I - A^{2^{n + 1} } } \circ \paren {I + A^{2^{n + 1} } }\)
\(\ds \) \(=\) \(\ds I + A^{2^{n + 1} } - A^{2^{n + 1} } + \paren {A^{2^{n + 1} } }^2\)
\(\ds \) \(=\) \(\ds I - A^{2 \cdot 2^{n + 1} }\)
\(\ds \) \(=\) \(\ds I - A^{2^{n + 2} }\)
\(\ds \) \(=\) \(\ds I - A^{2^{\paren {n + 1} + 1} }\)

$\Box$


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

We have that $\norm A < 1$.

Furthermore:

\(\ds \norm {I - A^{2^{n + 1} } - I}\) \(=\) \(\ds \norm { - A^{2^{n + 1} } }\)
\(\ds \) \(=\) \(\ds \norm {A^{2^{n + 1} } }\)
\(\ds \) \(\le\) \(\ds \norm A^{2^{n + 1} }\)
\(\ds \leadsto \ \ \) \(\ds \lim_{n \mathop \to \infty} \norm {I - A^{2^{n + 1} } - I}\) \(=\) \(\ds 0\)

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:

\(\ds \norm {\paren {I - A}^{-1} \circ \paren {I - A^{2^{n + 1} } } - \paren {I - A}^{-1} }\) \(=\) \(\ds \norm {\paren {I - A}^{-1} \circ \paren {I - A^{2^{n + 1} } - I } }\)
\(\ds \) \(\le\) \(\ds \norm {\paren {I - A}^{-1} } \norm {I - A^{2^{n + 1} } - I}\)
\(\ds \leadsto \ \ \) \(\ds \lim_{n \mathop \to \infty} \norm {\paren {I - A}^{-1} \circ \paren {I - A^{2^{n + 1} } } - \paren {I - A}^{-1} }\) \(\le\) \(\ds \lim_{n \mathop \to \infty} \paren {\norm {\paren {I - A}^{-1} } \norm {I - A^{2^{n + 1} } - I} }\)
\(\ds \) \(=\) \(\ds \norm {\paren {I - A}^{-1} } \lim_{n \mathop \to \infty} \norm {I - A^{2^{n + 1} } - I}\)
\(\ds \) \(=\) \(\ds 0\)

Since:

\(\ds \sequence {\paren {I - A}^{-1} \circ \paren {I - A^{2^{n + 1} } } }_{n \mathop \in \N}\) \(=\) \(\ds \sequence {\paren {I - A}^{-1} \circ \paren {I - A} \circ P_n}_{n \mathop \in \N}\)
\(\ds \) \(=\) \(\ds \sequence {P_n}_{n \mathop \in \N}\)

Thus, $\sequence {P_n}_{n \mathop \in \N}$ converges to $\paren {I - A}^{-1}$ in $\struct {\map {CL} X, \norm {\, \cdot \,} }$.

$\blacksquare$


Sources

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