Element of Unital Banach Algebra Close to Identity is Invertible

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $\struct {A, \norm \cdot}$ be a unital Banach algebra over $\Bbb F$ with identity element $\mathbf 1_A$.

Let $a \in A$ be such that:

$\norm {\mathbf 1_A - a} < 1$


Then $a$ is invertible with inverse element $a^{-1}$ satisfying:

$\ds \norm {a^{-1} } \le \frac 1 {1 - \norm {\mathbf 1_A - a} }$


Proof

Let:

$x = \mathbf 1_A - a$

From Bound on Norm of Power of Element in Normed Algebra, we have:

$\norm {x^n} \le \norm x^n$

for each $n \in \Z_{\ge 0}$.

Then we have:

$\ds \sum_{n \mathop = 0}^\infty \norm {x^n} \le \sum_{n \mathop = 0}^\infty \norm x^n$

Since $\norm x < 1$, we have:

$\ds \sum_{n \mathop = 0}^\infty \norm x^n$ converges

from Sum of Infinite Geometric Sequence.

So:

$\ds \sum_{n \mathop = 0}^\infty \norm {x^n}$ converges.

Since $A$ is a Banach space, we have:

$\ds \sum_{n \mathop = 0}^\infty x^n$ converges.

by Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach.

Let:

$\ds b = \sum_{n \mathop = 0}^\infty x^n$

We show that:

$\ds a b = \mathbf 1_A$

and:

$\ds b a = \mathbf 1_A$

Note that:

$a = \mathbf 1_A - x$

Note that for each $N \in \N$, we have:

\(\ds \paren {\mathbf 1_A - x} \sum_{n \mathop = 0}^N x^n\) \(=\) \(\ds \sum_{n \mathop = 0}^N \paren {\mathbf 1_A - x} x^n\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^N \paren {x^n - x^{n + 1} }\)
\(\ds \) \(=\) \(\ds \mathbf 1_A - x^{N + 1}\)

and similarly:

\(\ds \paren {\sum_{n \mathop = 0}^N x^n} \paren {\mathbf 1_A - x}\) \(=\) \(\ds \sum_{n \mathop = 0}^N x^n \paren {\mathbf 1_A - x}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^N \paren {x^n - x^{n + 1} }\)
\(\ds \) \(=\) \(\ds \mathbf 1_A - x^{N + 1}\)

We have:

\(\ds \norm {\mathbf 1_A - x^{N + 1} - \mathbf 1_A}\) \(=\) \(\ds \norm {x^{N + 1} }\)
\(\ds \) \(\le\) \(\ds \norm x^{N + 1}\)
\(\ds \) \(\to\) \(\ds 0\) since $\norm x < 1$

so that:

$\mathbf 1_A - x^{N + 1} \to \mathbf 1_A$

from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence.

So taking $N \to \infty$ in:

$\ds a \sum_{n \mathop = 0}^N x^n = \mathbf 1_A - x^{N + 1}$

we obtain:

$a b = \mathbf 1_A$

from Product Rule for Sequence in Normed Algebra.

Similarly taking $N \to \infty$ in:

$\ds \paren {\sum_{n \mathop = 0}^N x^n} a = \mathbf 1_A - x^{N + 1}$

we obtain:

$b a = \mathbf 1_A$

So $a$ is invertible with inverse element $b$.

It remains to show that:

$\ds \norm b \le \frac 1 {1 - \norm x}$

For each $N \in \N$, we have that:

\(\ds \norm {\sum_{n \mathop = 0}^N x^n}\) \(\le\) \(\ds \sum_{n \mathop = 0}^N \norm {x^n}\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(\le\) \(\ds \sum_{n \mathop = 0}^N \norm x^n\) Bound on Norm of Power of Element in Normed Algebra
\(\ds \) \(\le\) \(\ds \sum_{n \mathop = 0}^\infty \norm x^n\)
\(\ds \) \(=\) \(\ds \frac 1 {1 - \norm x}\) Sum of Infinite Geometric Sequence

Taking $N \to \infty$, we have:

$\ds \norm b \le \frac 1 {1 - \norm x} = \frac 1 {1 - \norm {\mathbf 1_A - a} }$

from Modulus of Limit: Normed Vector Space.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Element_of_Unital_Banach_Algebra_Close_to_Identity_is_Invertible&oldid=619909"