\section{Invertible Elements}
\label{section:invertible-elements}


\begin{definition}[Invertible]
\label{definition:banach-algebra-invertible}
    Let $A$ be a unital Banach algebra and $x \in A$, then $x$ is \textbf{invertible} if there exists $x^{-1} \in A$ such that $xx^{-1} = x^{-1}x = 1$. The set $G(A)$ denotes the group of all invertible elements in $A$. 
\end{definition}

\begin{lemma}[Neumann Series]
\label{lemma:neumann-series}
    Let $A$ be a unital banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
    \[
    x^{-1} = \sum_{n = 0}^\infty (1 - x)^n
    \]

    and $\normn{x^{-1}}_A \le (1 - \norm{x})_A^{-1}$. 
\end{lemma}
\begin{proof}
    Since $\norm{1 - x}_A < 1$, the series converges absolutely. Let $y = \sum_{n = 0}^\infty (1 - x)^n$, then
    \[
    (1 - x) \sum_{n = 0}^\infty (1 - x)^n = \sum_{n = 0}^\infty (1 - x)^n - 1 = \sum_{n = 0}^\infty (1 - x)^n (1 - x)
    \]

    so $(1 - x)y = y - 1 = y(1 - x)$, and $xy = yx = 1$. 
\end{proof}

\begin{proposition}
\label{proposition:banach-algebra-inverse}
    Let $A$ be a unital Banach algebra, then:
    \begin{enumerate}
        \item $G(A)$ is open. 
        \item For any $x \in G(A)$ and $y \in B_A(0, \normn{x^{-1}}_A^{-1})$, 
        \[
        (x - y)^{-1} = x^{-1}\sum_{n = 0}^\infty (yx^{-1})^n
        \]

        \item The map $G(A) \to G(A)$ defined by $x \mapsto x^{-1}$ is $C^\infty$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (2): For any $x \in G(A)$ and $y \in B(0, \normn{x^{-1}}_A^{-1})$, $(x - y) = (1 - yx^{-1})x$. By \autoref{lemma:neumann-series}, 
    \[
    (1 - yx^{-1})^{-1} = \sum_{n = 0}^\infty (yx^{-1})^n
    \]

    so
    \[
    (x - y)^{-1} = x^{-1}\sum_{n = 0}^\infty (yx^{-1})^n
    \]

    (3): Since the inversion map is locally a power series, it is $C^\infty$ by \autoref{theorem:termwise-differentiation}. 
\end{proof}

\begin{corollary}
\label{corollary:invertible-boundary-explode}
    Let $A$ be a unital Banach algebra, $x \in A \setminus G(A)$, and $r > 0$, then $\normn{y^{-1}}_A > 1/r$ for all $y \in B(x, r) \cap G(A)$. 
\end{corollary}
\begin{proof}
    Let $y \in B(x, r)$, then $B(y, \normn{y^{-1}}_A^{-1}) \subset G(A)$ by \autoref{proposition:banach-algebra-inverse}. Since $x \not\in G(A)$, $r > \norm{x - y}_A \ge \normn{y^{-1}}_A^{-1}$, and $1/r < \normn{y^{-1}}_A$. 
\end{proof}


\begin{proposition}
\label{proposition:swap-invertible}
    Let $A$ be a unital Banach algebra and $x, y \in A$, then $1 - xy \in G(A)$ if and only if $1 - yx \in G(A)$. 
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition I.3.4]{Zhu}}}. ]
    If $1 - xy \in G(A)$, then
    \begin{align*}
        (1 - yx)[y(1 - xy)^{-1}x + 1] &= y(1 - xy)^{-1}x + 1 \\
        &-yxy(1 - xy)^{-1}x - yx \\
        &= y[(1 - xy)^{-1} - xy(1 - xy)^{-1}]x + 1 - yx \\
        &= yx + 1 - yx = 1
    \end{align*}
    
    Similarly, 
    \begin{align*}
        [y(1 - xy)^{-1}x + 1](1 - yx) &=  y(1 - xy)^{-1}x + 1 \\
        &-y(1 - xy)^{-1}xyx - yx \\
        &= y[(1 - xy)^{-1} - (1 - xy)^{-1}xy]y + 1 - yx \\
        &= yx + 1 - yx = 1
    \end{align*}
    
\end{proof}
