Element of Unital Banach Algebra on Boundary of Group of Units of Subalgebra is Not Invertible in Algebra/Lemma
Lemma
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $\map G A$ be the group of units of $A$.
Let $x \in \partial \map G A$, where $\partial \map G A$ is the topological boundary of $\map G A$.
Then there exists a sequence $\sequence {z_n}_{n \in \N}$ in $A$ such that $\norm {z_n} = 1$ for each $n \in \N$, and:
- $z_n x \to 0$ as $n \to \infty$
and:
- $x z_n \to 0$ as $n \to \infty$.
Proof
From Group of Units in Unital Banach Algebra is Open, we have:
- $\partial \map G A = \map G A^- \setminus \map G A$
So if $x \in \partial \map G A$, then $x \in A \setminus \map G A$ and there exists a sequence $\sequence {x_n}_{n \in \N}$ with $x_n \in \map G A$ for each $n \in \N$, such that:
- $x_n \to x$ as $n \to \infty$.
Let:
- $\ds z_n = \frac {x_n^{-1} } {\norm {x_n^{-1} } }$
We have:
- $\ds \norm {z_n} = \norm {\frac {x_n^{-1} } {\norm {x_n^{-1} } } } = \frac {\norm {x_n^{-1} } } {\norm {x_n^{-1} } } = 1$
from Norm Axiom $\text N 2$: Positive Homogeneity, for each $n \in \N$.
Now, we have:
\(\ds \norm {z_n x}\) | \(=\) | \(\ds \norm {z_n \paren {x - x_n} + x_n z_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {z_n \paren {x - x_n} } + \norm {x_n z_n}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x - x_n} + \frac 1 {\norm {x_n^{-1} } }\) | Definition of Algebra Norm, Definition of Unital Banach Algebra |
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We know by hypothesis that $\norm {x - x_n} \to 0$, while from Norm of Inverse of Sequence of Invertible Elements Converging to Non-Invertible Element in Unital Banach Algebra, we have:
- $\ds \frac 1 {\norm {x_n^{-1} } } \to 0$
So, we have:
- $z_n x \to 0$ as $n \to \infty$.
Similarly, we have:
\(\ds \norm {x z_n}\) | \(=\) | \(\ds \norm {\paren {x - x_n} z_n + x_n z_n}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {\paren {x - x_n} z_n} + \norm {x_n z_n}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x - x_n} + \frac 1 {\norm {x_n^{-1} } }\) | Definition of Algebra Norm, Definition of Unital Banach Algebra |
So $\sequence {z_n}_{n \in \N}$ satisfies the conditions given in the theorem.
$\blacksquare$