Element of Unital Banach Algebra on Boundary of Group of Units of Subalgebra is Not Invertible in Algebra

Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.

Let $\map G A$ be the group of units of $A$.

Let $B$ be a closed subalgebra of $A$.

Let $\map G B$ be the group of units of $A$.

Let $x \in \partial \map G B$, where $\partial \map G B$ is the topological boundary of $\map G B$.

Then $x$ is not invertible in $A$.

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 $n \to \infty$.

$\Box$

Aiming for a contradiction, suppose that there exists $y \in A$ such that $x y = 1$.

From the lemma, 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 $n \to \infty$

$z_n x y \to 0$ as $n \to \infty$.

Since $x y = 1$, this gives $z_n \to 0$ as $n \to \infty$.

From Modulus of Limit: Normed Vector Space, we then have $\norm {z_n} \to 0$ as $n \to \infty$.

But $\norm {z_n} = 1$ for each $n \in \N$, so this is impossible.

So we have a contradiction and there exists no such $y \in A$.

$\blacksquare$