Proposition 29.2.4.label Let $A$ be a unital Banach algebra and $x, y \in A$, then $1 - xy \in G(A)$ if and only if $1 - yx \in G(A)$.
Proof, [Proposition I.3.4, Zhu93]. 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*}
$\square$
Post a Comment