Group of Units in Unital Banach Algebra is Open
Theorem
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $\map G A$ be the group of units of $A$.
Then $\map G A$ is open in $A$.
Proof
Let $x \in \map G A$.
We find an open neighborhood of $x$ contained in $\map G A$.
Clearly $x^{-1} \ne \mathbf 0_A$, so $\norm {x^{-1} } > 0$.
We have, for $y \in A$:
\(\ds \norm {1 - x^{-1} y}\) | \(=\) | \(\ds \norm {x^{-1} \paren {x - y} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x^{-1} } \norm {x - y}\) |
So for $y \in A$ satisfying:
- $\ds \norm {x - y} < \frac 1 {\norm {x^{-1} } }$
we have:
- $\norm {1 - x^{-1} y} < 1$
From Element of Unital Banach Algebra Close to Identity is Invertible, we have that $x^{-1} y$ is invertible.
That is, $x^{-1} y \in \map G A$.
Since $\map G A$ is a group, from Group of Units is Group, we have:
- $y = x \paren {x^{-1} y} \in \map G A$
So we have:
- $\map {B_{\norm {x^{-1} }^{-1} } } x \subseteq \map G A$
where $\map {B_{\norm {x^{-1} }^{-1} } } x$ is the open ball centered at $x$ with radius $\norm {x^{-1} }^{-1}$.
Since $x \in \map G A$ was arbitrary, we have that $\map G A$ is open.
$\blacksquare$
Sources
- 2011: Graham R. Allan and H. Garth Dales: Introduction to Banach Spaces and Algebras ... (previous) ... (next): $4.4$: The group of units