# Permanently Singular Elements in a Banach Algebra

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

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

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

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

Proposition 1: Let $X$ be a normed algebra with unit. If $x$ is a left topological divisor of zero or a right topological divisor of zero then $x$ is permanently singular. |

**Proof:**Let $x \in \mathrm{Sing}(X)$ be a left topological divisor of zero. Suppose instead that there exists an normed algebra extension $W$ of $X$ such that $x \not \in \mathrm{Sing}(W)$. Then $x$ is invertible in $W$. So there exists a $y \in W$ such that $xy = yx = 1$.

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

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

- Since $W$ is a normed algebra extension of $X$, 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 $\| x \|_X \leq M \| x \|_W$ for all $x \in X$. Since $(y_n)$ is a sequence in $X$, 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 $x$ is not permanently singular is false. Thus, $x$ is permanently singular.

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

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

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