Element of Unital Banach Algebra Close to Identity is Invertible

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:


 * $\ds \norm {a^{-1} } \le \frac 1 {1 - \norm x}$

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:


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

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:

and similarly:

We have:

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 Limits 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:


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