# Proof involving invertible elements of Banach algebra

I want to prove for a unital Banach algebra $\mathcal{A}$, it follows that if $\|a-b \| < \frac{1}{\|a^{-1} \|}$ then $b \in \mathcal{A}^{-1}$ (where $\mathcal{A}^{-1}$ is the subset of invertible elements of $\mathcal{A}$).

I can prove that this implication holds for the strict inequality i.e. $\| a - b \| < \frac{1}{\| a^{-1} \|} \implies$ $b \in \mathcal{A}^{-1}$:

\begin{align} &~~~~~~~~\| (a-b)a^{-1} \| \leq \| a-b \| \| a^{-1} \| \leq 1 \\& \therefore~~~~~ \|aa^{-1} - ba^{-1} \| \leq \| a-b \| \| a^{-1} \| \leq 1 \\& \therefore~~~~~ \| 1 - ba^{-1} \| \leq 1 \\& \therefore~~~~~ \| ba^{-1}-1 \| \leq 1 \end{align}

We have that $ba^{-1} \in B_{1}(1)$. Let $z = 1-ba^{-1} \implies \| z \| < 1$. Let $c = \sum\limits_{k=0}^{\infty}z^{k} = \sum\limits_{k=0}^{\infty}(1-ba^{-1})^{k}$. We can show that this geometric series converges using the completeness of $\mathcal{A}$. We then obtain: \begin{align} (ba^{-1})c = (1-z)c = \sum\limits_{k=0}^{\infty}(1-ba^{-1})^{k}-\sum\limits_{k=0}^{\infty}(1-ba^{-1})^{k+1} = 1. \end{align} Similarly, $c(ba^{-1}) = 1 \implies ba^{-1} \in \mathcal{A}^{-1}$ $\implies b \in \mathcal{A}^{-1}$.

I'm having difficulty proving this for the case $\|a-b \| = \frac{1}{\| a^{-1} \|} \implies b \in \mathcal{A}^{-1}$, since in that case you cannot use the geometric series in the way that I used it above.

Thanks for any assistance.

Let $\mathcal{A}=C[0,1]$, $a\equiv 1$, and let $b$ be $1$ on $[0,1/2]$, taper linearly to $0$ on $[1/2,3/4]$ and be $0$ on $[3/4,1]$. Then $\|a\|=\|a^{-1}\|=1$ and $$\|a-b\| = 1 = \frac{1}{\|a^{-1}\|}.$$ However, $b$ is not invertible.
• More simply, just let $\mathcal{A}=\mathbb{C}$ and $b=0$. – Eric Wofsey Sep 9 '15 at 21:24