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

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

(1)

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

We also proved that the spectral radius of a point $a \in \mathfrak{A}$ can be obtained by the formula:

(2)

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

Now if $\mathfrak{A}$ is a Banach algebra with unit, $a \in \mathfrak{A}$, and $r(a) < 1$, we will see by the following theorem that $1 - a$ is an invertible element of $\mathfrak{A}$.

Theorem 1: Let $\mathfrak{A}$ be a Banach algebra with unit and let $a \in \mathfrak{A}$. If $r(a) < 1$ then $1 - a$ is invertible and furthermore $\displaystyle{(1 - a)^{-1} = 1 + \sum_{n=1}^{\infty} a^n}$.

• Proof: Suppose that $r(a) < 1$. Let $t$ be such that $r(a) < t < 1$. Now since $\displaystyle{r(a) = \lim_{n \to \infty} \| a^n \|^{1/n}}$ for this number $t > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\| a^n \|^{1/n} \leq t$, or equivalently, if $n \geq N$ then:

(3)

\begin{align} \quad \| a^n \| \leq t^n \end{align}

• Therefore we have that:

(4)

\begin{align} \quad \left \| 1 + \sum_{n=1}^{\infty} a^n \right \| \leq \| 1 \| + \sum_{n=1}^{\infty} \| a^n \| = \| 1 \| + \sum_{n=1}^{N-1} \| a^n \| + \sum_{n=N}^{\infty} \| a^n \| \leq \| 1 \| + \sum_{n=1}^{N-1} \| a^n \| + \underbrace{\sum_{n=N}^{\infty} t^n}_{\: < \infty \: \mathrm{since \:} |t| < 1} < \infty \end{align}

• So the series $\displaystyle{1 + \sum_{n=1}^{\infty} a^n}$ converges, say:

(5)

\begin{align} \quad y = 1 + \sum_{n=1}^{\infty} a^n \end{align}

• Since $\mathfrak{A}$ is a Banach algebra (and hence a Banach space), we see that $y \in \mathfrak{A}$. For each $n \in \mathbb{N}$ let:

(6)

\begin{align} \quad s_n = 1 + a + a^2 + ... + a^{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 - a)s_n = (1 - a) (1 + a + a^2 + ... + a^{n-1}) = 1 - a^n \quad \mathrm{and} \quad s_n(1 - a) = (1 + a + a^2 + ... + a^{n-1})(1 - a) = 1 - a^n \quad (*) \end{align}

• Since multiplication in a Banach algebra is continuous, we have that $((1 - a)s_n)$ converges to $(1 - a)y$, $(s_n(1 - a))$ converges to $y(1 - a)$, and lastly, $1 - a^n$ converges to $1$ since:

(8)

\begin{align} \quad \lim_{n \to \infty} \| [1 - a^n] - 1 \| \leq \lim_{n \to \infty} \| a^n \| \overset{(\dagger)}= 0 \end{align}

• (Where the equality at $(\dagger)$ comes from the fact that $\displaystyle{\sum_{n=1}^{\infty} a^n < \infty}$ and so $\displaystyle{\lim_{n \to \infty} \| a^n \| = 1}$). Thus by the equality at $(*)$ we see that:

(9)

\begin{align} \quad (1 - a)y = y(1 - a) = 1 \end{align}

• So $(1 - a) \in \mathrm{Inv}(\mathfrak{A})$ and furthermore $(1 - a)^{-1} = y$, that is:

(10)

\begin{align} \quad (1 - a)^{-1} = 1 + \sum_{n=1}^{\infty} a^n \quad \blacksquare \end{align}

Corollary 2: Let $\mathfrak{A}$ be a Banach algebra with unit and let $a \in \mathfrak{A}$. If $\| 1 - a \| < 1$ then $a$ is invertible.

• Proof: Let $a \in \mathfrak{A}$ be such that $\| 1 - a \| < 1$. Then:

(11)

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

• Therefore by Theorem 1 we have that $a = 1 - (1 - a)$ is invertible. $\blacksquare$