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, for each $n \in \N$.

Now, we have:

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:

So $\sequence {z_n}_{n \in \N}$ satisfies the conditions given in the theorem.