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.
 * $(3) \quad$ $\ds \paren {I - T}^{-1} = \sum_{k \mathop = 0}^\infty T^k$
 * $(4) \quad$ $\norm {\paren {I - T}^{-1} } \le \paren {1 - \norm T}^{-1}$

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:

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:

Note that we have:


 * $\ds \frac {\norm T^{m + 1} } {1 - \norm T} < \epsilon$

we have:


 * $\norm T^{m + 1} < \epsilon \paren {1 - \norm T}$

Taking logarithms, this holds :


 * $\paren {m + 1} \ln \norm T < \map \ln {\epsilon \paren {1 - \norm T} }$

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:

We have:

so:


 * $\ds \lim_{n \to \infty} R_n \paren {I - T} = I$

so:


 * $R \paren {I - T} = I$

Similarly, we have:

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 $(1)$.

Proof of $(2)$
Note that since $R$ is bounded, we immediately have that:


 * $\paren {I - T}^{-1}$ is bounded.

Proof of $(3)$
Since:


 * $\ds R = \sum_{k \mathop = 0}^\infty T^k$

and:


 * $R = \paren {I - T}^{-1}$

we obtain $(3)$ immediately.

Proof of $(4)$
We have:

So, from Limits Preserve Inequalities:


 * $\norm R \le \dfrac 1 {1 - \norm T}$

giving the bound required by $(4)$.