# Permanently Singular Elements in a Banach Algebra with Unit

Recall from the Boundary Points of Inv(A) are Topological Divisors of Zero in a Banach Algebra with Unit page that if $\mathfrak{A}$ is a Banach algebra with unit and $a \in \partial (\mathrm{Inv}(\mathfrak{A}))$ then $a$ is a topological divisor of zero.

Given any normed algebra with unit $\mathfrak{A}$, we can consider a point $a \in \mathrm{Sing}(\mathfrak{A})$. If $\mathfrak{B}$ is a normed algebra extension of $\mathfrak{A}$ then the point $a$ may or may not be a singular element of $\mathfrak{A}$. We give a special name to a point $a$ which is singular in every normed algebra extension $\mathfrak{B}$ of $\mathfrak{A}$.

Definition: Let $\mathfrak{A}$ be a normed algebra with unit. A point $a \in \mathrm{Sing}(\mathfrak{A})$ is said to be Permanently Singular if for every normed algebra extension $\mathfrak{B}$ of $\mathfrak{A}$ we have that $a \in \mathrm{Sing}(\mathfrak{B})$. |

*If $\mathfrak{A}$ is a normed algebra with unit then another normed algebra with unit $\mathfrak{B}$ is said to be an extension of $\mathfrak{A}$ if: $\mathfrak{A} \subseteq \mathfrak{B}$; the norm on $\mathfrak{B}$ when restricted to $\mathfrak{A}$ is norm equivalent to the norm on $\mathfrak{A}$; and the unit of $\mathfrak{B}$ is equal to the unit of $\mathfrak{A}$.*

Proposition 1: Let $\mathfrak{A}$ be a normed algebra with unit. If $a$ is a left topological divisor of zero or a right topological divisor of zero then $a$ is permanently singular. |

**Proof:**Let $a \in \mathrm{Sing}(\mathfrak{A})$ be a left topological divisor of zero. Suppose instead that there exists an normed algebra extension $\mathfrak{B}$ of $\mathfrak{A}$ such that $a \not \in \mathrm{Sing}(\mathfrak{B})$. Then $a$ is invertible in $\mathfrak{B}$. So there exists a $y \in \mathfrak{B}$ such that $ay = ya = 1$.

- Since $a$ is a left topological divisor of zero there exists a sequence of points $(y_n)$ with $\| y_n \|_{\mathfrak{A}} = 1$ for all $n \in \mathbb{N}$ such that:

- Then for every $n \in \mathbb{N}$:

- Since $\mathfrak{B}$ is a normed algebra extension of $\mathfrak{A}$, we have that $\| \cdot \|_X$ is norm equivalent to $\| \cdot \|_W$ when restricted to $X$. In particular, there exists a number $M > 0$ such that $\| x \|_{\mathfrak{A}} \leq M \| x \|_{\mathfrak{B}}$ for all $x \in \mathfrak{A}$. Since $(y_n)$ is a sequence in $\mathfrak{A}$, the above inequality tells us that for every $n \in \mathbb{N}$:

- Taking the limit of both sides of the inequality at $n \to \infty$ gives us that $1 \leq 0$, a contradiction. Therefore the assumption that $a$ is not permanently singular is false. Thus, $a$ is permanently singular.

- An analogous argument shows that if $a$ is a right topological divisor of zero then $a$ is also permanently singular. $\blacksquare$

Corollary 2: Let $\mathfrak{A}$ be a Banach algebra with unit. If $a \in \partial (\mathrm{Inv}(\mathfrak{A}))$ then $a$ is permanently singular. |

**Proof:**Let $a \in \partial (\mathrm{Inv}(\mathfrak{A}))$. From the Theorem mentioned on the top of the page, $a$ is thus a topological divisor of zero. By Proposition 1, $a$ is permanently singular. $\blacksquare$