Quasi-Invertibility of x When r(x) < 1 in a Banach Algebra

# Quasi-Invertibility of x When r(x) < 1 in a Banach Algebra

Theorem 1: Let $X$ be a Banach algebra and let $x \in X$. If $r(x) < 1$ then $x$ is quasi-invertible and the quasi-inverse of $x$ is $\displaystyle{x^0 = - \sum_{n=1}^{\infty} x^n}$. |

*Note that we do not require $X$ to have a unit.*

**Proof:**Since $X$ is a Banach algebra, its unitization $X + \mathbf{F}$ is also a Banach algebra. Consider the point $(x, 0) \in X + \mathbf{F}$. We have that:

\begin{align} \quad r((x, 0)) &= \inf \left \{ \| (x, 0)^n \|^{1/n} : n \in \mathbb{N} \right \} \\ &= \inf \left \{ \| (x^n, 0) \|^{1/n} : n \in \mathbb{N} \right \} \\ &= \inf \left \{ (\| x^n \| + |0|)^{1/n} : n \in \mathbb{N} \right \} \\ &= \inf \left \{ \| x^n \|^{1/n} : n \in \mathbb{N} \right \} \\ &= r(x) \\ & < 1 \end{align}

- So by the theorem on the Invertibility of 1 - x When r(x) < 1 in a Banach Algebra page, since $X + \mathbf{F}$ is a Banach algebra with unit $(0, 1)$, and $r((x, 0)) < 1$ we have that since $(0, 1) - (x, 0)$ is invertible and the inverse is given by:

\begin{align} \quad [(0, 1) - (x, 0)]^{-1} &= (0, 1) + \sum_{n=1}^{\infty} (x, 0)^n \\ &= (0, 1) - \left ( - \sum_{n=1}^{\infty} (x, 0)^n \right ) \\ &= (0, 1) - \left ( - \sum_{n=1}^{\infty} (x^n, 0) \right ) \\ &= (0, 1) - \left (- \sum_{n=1}^{\infty} x^n, 0 \right ) \end{align}

- (The sum can be pulled inside the brackets because addition in $X + \mathbf{F}$ is continuous as $X + \mathbf{F}$ is a normed space and hence a topological vector space).

- But from one of the results on the Basic Theorems Regarding Quasi-Invertible Elements in an Algebra page we have that this implies that $x$ is quasi-invertible with quasi-inverse:

\begin{align} \quad x^0 &= - \sum_{n=1}^{\infty} x^n \quad \blacksquare \end{align}