33.2 Invertible Elements
Definition 33.2.1 (Invertible).label Let $A$ be a unital Banach algebra and $x \in A$, then $x$ is 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 $A$.
Lemma 33.2.2 (Neumann Series).label Let $A$ be a unital banach algebra and $x \in B_{A}(1, 1)$, then $x \in G(A)$ with
and $\normn{x^{-1}}_{A} \le (1 - \norm{x})_{A}^{-1}$.
Proof. Since $\norm{1 - x}_{A} < 1$, the series converges absolutely. Let $y = \sum_{n = 0}^{\infty} (1 - x)^{n}$, then
so $(1 - x)y = y - 1 = y(1 - x)$, and $xy = yx = 1$.$\square$
Proposition 33.2.3.label Let $A$ be a unital Banach algebra, then:
- (1)
$G(A)$ is open.
- (2)
For any $x \in G(A)$ and $y \in B_{A}(0, \normn{x^{-1}}_{A}^{-1})$,
\[(x - y)^{-1}= x^{-1}\sum_{n = 0}^{\infty} (yx^{-1})^{n}\] - (3)
The map $G(A) \to G(A)$ defined by $x \mapsto x^{-1}$ is $C^{\infty}$.
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 Lemma 33.2.2,
so
(3): Since the inversion map is locally a power series, it is $C^{\infty}$ by Theorem 30.7.3.$\square$
Proposition 33.2.4.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
Similarly,
$\square$
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