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

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

Recall from The Spectral Radius of a Point in a Normed Algebra page that if $X$ is a normed algebra and $x \in X$ then the spectral radius of $x$ is defined to be:

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

We also proved that the spectral radius of a point $x \in X$ can be obtained by the formula:

(2)
\begin{align} \quad r(x) = \lim_{n \to \infty} \| x^n \|^{1/n} \end{align}

Now if $X$ is a Banach algebra, $x \in X$, and $r(x) < 1$, we will see by the following theorem that $1 - x$ is an invertible element of $X$.

 Theorem 1: Let $X$ be a Banach algebra with unit and let $x \in X$. If $r(x) < 1$ then $1 - x$ is invertible and furthermore $\displaystyle{(1 - x)^{-1} = 1 + \sum_{n=1}^{\infty} x^n}$.
• Proof: Suppose that $r(x) < 1$. Let $t$ be such that $r(x) < t < 1$. Now since $\displaystyle{r(x) = \lim_{n \to \infty} \| x^n \|^{1/n}}$ for this number $t > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\| x^n \|^{1/n} \leq t$, or equivalently, if $n \geq N$ then:
(3)
\begin{align} \quad \| x^n \| \leq t^n \end{align}
• Therefore we have that:
(4)
\begin{align} \quad \left \| 1 + \sum_{n=1}^{\infty} x^n \right \| \leq \| 1 \| + \sum_{n=1}^{\infty} \| x^n \| = \| 1 \| + \sum_{n=1}^{N-1} \| x^n \| + \sum_{n=N}^{\infty} \| x^n \| \leq \| 1 \| + \sum_{n=1}^{N-1} \| x^n \| + \underbrace{\sum_{n=N}^{\infty} t^n}_{\: < \infty \: \mathrm{since \:} |t| < 1} < \infty \end{align}
• So the series $\displaystyle{1 + \sum_{n=1}^{\infty} x^n}$ converges, say:
(5)
\begin{align} \quad y = 1 + \sum_{n=1}^{\infty} x^n \end{align}
• Since $X$ is a Banach algebra (and hence a Banach space), we see that $y \in X$. For each $n \in \mathbb{N}$ let:
(6)
\begin{align} \quad s_n = 1 + x + x^1 + ... + x^{n-1} \end{align}
• That is, $(s_n)$ is the sequence of partial sums of $y$. Then $(s_n)$ norm converges to $y$. Now observe that for each $n \in \mathbb{N}$ we have that:
(7)
\begin{align} \quad \quad (1 - x)s_n = (1 - x) (1 + x + x^1 + ... + x^{n-1}) = 1 - x^n \quad \mathrm{and} \quad s_n(1 - x) = (1 + x + x^1 + ... + x^{n-1})(1 - x) = 1 - x^n \quad (*) \end{align}
• Since multiplication in a Banach algebra is continuous, we have that $((1 - x)s_n)$ converges to $(1 - x)y$, $(s_n(1 - x))$ converges to $y(1 - x)$, and lastly, $1 - x^n$ converges to $1$ since:
(8)
\begin{align} \quad \lim_{n \to \infty} \| [1 - x^n] - 1 \| \leq \lim_{n \to \infty} \| x^n \| \overset{(\dagger)}= 0 \end{align}
• (Where the equality at $(\dagger)$ comes from the fact that $\displaystyle{\sum_{n=1}^{\infty} x^n < \infty}$ and so $\displaystyle{\lim_{n \to \infty} \| x^n \| = 1}$). Thus by the equality at $(*)$ we see that:
(9)
\begin{align} \quad (1 - x)y = y(1 - x) = 1 \end{align}
• So $(1 - x) \in \mathrm{Inv}(X)$ and furthermore $(1 - x)^{-1} = y$, that is:
(10)
\begin{align} \quad (1 - x)^{-1} = 1 + \sum_{n=1}^{\infty} x^n \quad \blacksquare \end{align}
 Corollary 2: Let $X$ be a Banach algebra with unit and let $x \in X$. If $\| 1 - x \| < 1$ then $x$ is invertible.
• Proof: Let $x \in X$ be such that $\| 1 - x \| < 1$. Then:
(11)
\begin{align} \quad r(1 - x) = \inf \left \{ \| (1 - x)^n \|^{1/n} : n \in \mathbb{N} \right \} \leq \| (1 - x)^1 \|^1 = \| 1 - x \| < 1 \end{align}
• Therefore by theorem 1 we have that $x = 1 - (1 - x)$ is invertible. $\blacksquare$