User:Caliburn/s/fa/2
Theorem
Let $\struct {X, \norm \cdot_X}$ be a Banach space.
Let $\norm \cdot$ be the norm of a bounded linear transformation.
Let $\map B {X, X}$ be the space of bounded linear transformations $X \to X$.
Let $T \in \map B {X, X}$ be a bounded linear operator with $\norm T < 1$.
Then:
- $(1) \quad$ $I - T$ is invertible with inverse $\paren {I - T}^{-1}$
- $(2) \quad$ $\paren {I - T}^{-1}$ is bounded.
- $(3) \quad$ $\ds \paren {I - T}^{-1} = \sum_{k \mathop = 0}^\infty T^k$
- $(4) \quad$ $\norm {\paren {I - T}^{-1} } \le \paren {1 - \norm T}^{-1}$
Corollary
Let $A : X \to X$ be an invertible bounded linear operator with bounded inverse $A^{-1} : X \to X$.
Let $B : X \to X$ be an invertible bounded linear operator with $\norm B \norm {A^{-1} } < 1$.
Then:
- $(1) \quad$ $A + B$ is invertible with inverse $\paren {A + B}^{-1}$
- $(2) \quad$ $\paren {A + B}^{-1}$ is bounded.
- $(3) \quad$ $\norm {\paren {A + B}^{-1} } \le \norm {A^{-1} } \paren {1 - \norm {A^{-1} } \norm B}^{-1}$
Proof
Define the sequence $\sequence {R_n}_{n \in \N}$ by:
- $\ds R_n = \sum_{k \mathop = 0}^n T^k$
for each $n \in \N$.
We first aim to show that $\sequence {R_n}_{n \in \N}$ is Cauchy.
Let $n, m$ be natural numbers with $n > m$.
Let $\epsilon$ be a positive real number.
We have:
\(\ds \norm {R_n - R_m}\) | \(=\) | \(\ds \norm {\sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^m T^k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\sum_{k \mathop = m + 1}^n T^k + \sum_{k \mathop = 0}^m T^k - \sum_{k \mathop = 0}^m T^k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\sum_{k \mathop = m + 1}^n T^k}\) |
We have (need link):
- $\norm {T^k} \le \norm T^k$
for each $k \in \N$.
So, using the triangle inequality part of Norm on Bounded Linear Transformation is Norm, we have:
\(\ds \norm {\sum_{k \mathop = m + 1}^n T^k}\) | \(\le\) | \(\ds \sum_{k \mathop = m + 1}^n \norm {T^k}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{k \mathop = m + 1}^n \norm T^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \norm T^k - \sum_{k \mathop = 0}^m \norm T^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - \norm T^{n + 1} } {1 - \norm T} - \frac {1 - \norm T^{m + 1} } {1 - \norm T}\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\norm T^{m + 1} - \norm T^{n + 1} } {1 - \norm T}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac {\norm T^{m + 1} } {1 - \norm T}\) |
Note that we have:
- $\ds \frac {\norm T^{m + 1} } {1 - \norm T} < \epsilon$
if and only if we have:
- $\norm T^{m + 1} < \epsilon \paren {1 - \norm T}$
Taking logarithms, this holds if and only if:
- $\paren {m + 1} \ln \norm T < \map \ln {\epsilon \paren {1 - \norm T} }$
Since:
- $\norm T < 1$
we have:
- $\ds m > \frac {\map \ln {\epsilon \paren {1 - \norm T} } } {\norm T} - 1$
Set:
- $\ds N = \frac {\map \ln {\epsilon \paren {1 - \norm T} } } {\norm T} - 1$
Then, for $n > m > N$, we have:
- $\norm {R_n - R_m} < \epsilon$
So, we have that:
- $\sequence {R_n}_{n \in \N}$ is Cauchy.
From Space of Bounded Linear Transformations is Banach Space, we have that:
- $\struct {\map B {X, X}, \norm \cdot}$ is a Banach space.
So:
- $\sequence {R_n}_{n \in \N}$ converges in $\map B {X, X}$.
Let:
- $\ds R = \lim_{n \mathop \to \infty} R_n$
We can see that:
- $\ds R = \sum_{k \mathop = 0}^\infty T^k$
We now show that:
- $R = \paren {I - T}^{-1}$
and that $R$ satisfies the desired properties.
Proof of $(1)$
We have:
\(\ds R_n \paren {I - T}\) | \(=\) | \(\ds \paren {\sum_{k \mathop = 0}^n T^k} \paren {I - T}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^n T^{k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds I - T^{n + 1} + \sum_{k \mathop = 1}^n T^k - \sum_{k \mathop = 0}^{n - 1} T^{k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds I - T^{n + 1} + \sum_{k \mathop = 1}^n T^k - \sum_{k \mathop = 1}^n T^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds I - T^{n + 1}\) |
We have:
\(\ds \norm {R_n \paren {I - T} - I}\) | \(=\) | \(\ds \norm {T^{n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm T^{n + 1}\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 0\) |
so:
- $\ds \lim_{n \to \infty} R_n \paren {I - T} = I$
so:
- $R \paren {I - T} = I$
Similarly, we have:
\(\ds \paren {I - T} R_n\) | \(=\) | \(\ds \paren {I - T} \sum_{k \mathop = 0}^n T^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n T^k - \sum_{k \mathop = 0}^n T^{k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds I - T^{n + 1}\) |
So, taking $n \to \infty$ we have:
- $\paren {I - T} R = I$
So $I - T$ is invertible with:
- $R = \paren {I - T}^{-1}$
so we have $(1)$.
$\Box$
Proof of $(2)$
Note that since $R$ is bounded, we immediately have that:
- $\paren {I - T}^{-1}$ is bounded.
$\Box$
Proof of $(3)$
Since:
- $\ds R = \sum_{k \mathop = 0}^\infty T^k$
and:
- $R = \paren {I - T}^{-1}$
we obtain $(3)$ immediately.
$\Box$
Proof of $(4)$
We have:
\(\ds \norm {R_n}\) | \(=\) | \(\ds \norm {\sum_{k \mathop = 0}^n T^k}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{k \mathop = 0}^n \norm {T^k}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{k \mathop = 0}^n \norm T^k\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{k \mathop = 0}^\infty \norm T^k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {1 - \norm T}\) |
So, from Limits Preserve Inequalities:
- $\norm R \le \dfrac 1 {1 - \norm T}$
giving the bound required by $(4)$.