Added basics of Banach algebras.
This commit is contained in:
@@ -4,15 +4,17 @@
|
||||
|
||||
\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 collection of all invertible elements in $A$.
|
||||
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}
|
||||
\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
|
||||
@@ -50,3 +52,27 @@
|
||||
(3): Since the inversion map is locally a power series, it is $C^\infty$ by \autoref{theorem:termwise-differentiation}.
|
||||
\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}
|
||||
|
||||
|
||||
|
||||
Reference in New Issue
Block a user