Topological Divisors of Zero in a Normed Algebra

# Topological Divisors of Zero in a Normed Algebra

Definition: Let $X$ be a normed algebra.1. A point $x \in X$ is said to be a Left Topological Divisor of Zero if $\inf \{ \| xy \| : \| y \| = 1 \} = 0$.2. A point $x \in X$ is said to be a Right Topological Divisor of Zero if $\inf \{ \| yx \| : \| y \| = 1 \} = 0$.3. A point $x \in X$ is said to be a (Joint) Topological Divisor of Zero if $\inf \{ \| xy \| + \| yx \| : \| y \| = 1 \} = 0$. |

The following proposition tells us that if $x$ is a topological divisor of zero then it must also be a left topological divisor of zero and a right topological divisor of zero.

Proposition 1: Let $X$ be a normed algebra. If $x \in X$ is a topological divisor of zero then $x$ is both a left topological divisor of zero and a right topological divisor of zero. |

**Proof:**This follows immediately from the definition above. Indeed, if $x$ is a topological divisor of zero then $\inf \{ \| xy \| + \| yx \| : \| y \| = 1 \} = 0$. If $x$ is NOT a left topological divisor of zero then $\inf \{ \| xy \| : \| y \| = 1 \} > 0$ and since the sum $\| xy \| + \| yx \| \geq 0$ for all $y$ with $\| y \| = 1$ we see that $\inf \{ \| xy \| + \| yx \| : \| y \| = 1 \} > 0$, a contradiction, and so $x$ must be a left topological divisor of zero.

- A similar argument shows that $x$ is also a right topological divisor of zero. $\blacksquare$

Proposition 2: Let $X$ be a normed algebra. Then $x \in X$ is a topological divisor of zero if and only if there exists a sequence $(y_n)$ with $\| y_n \| = 1$ for all $n \in \mathbb{N}$ such that $\displaystyle{\lim_{n \to \infty} \| xy_n \| = 0}$ and $\displaystyle{\lim_{n \to \infty} \| y_nx \| = 0}$. |

**Proof:**$\Rightarrow$ Suppose that $x$ is a topological divisor of zero. Then by proposition 1 it is both a left topological divisor of zero and a right topological divisor of zero.

- Since $x$ is a left topological divisor of zero we have that $\inf \{ \| xy \| : \| y \| = 1 \} = 0$. So for every $n \in \mathbb{N}$ there exists a $z_n \in X$ with $\| z_n \| = 1$ such that $\| xz_n \| \leq n$. Therefore:

\begin{align} \quad \lim_{n \to \infty} \| xz_n \| \leq \lim_{n \to \infty} \| x \| \| z_n \| \leq \lim_{n \to \infty} \frac{\| x \|}{n} = 0 \end{align}

- An analogous argument shows that a sequence $(z_n')$ with $\| z_n' \| = 1$ for all $n \in \mathbb{N}$ exists with:

\begin{align} \quad \lim_{n \to \infty} \| z_n'x \| = 0 \end{align}

- Take $(y_n)$ to be the sequence whose odd terms are obtained from $(z_n)$ and whose even terms are obtained from $(z_n')$ so that $(y_n)$ satisfies the remaining conditions of this direction of the proof.

- $\Leftarrow$ Suppose that there exists a sequence $(y_n)$ with $\| y_n \| = 1$ for all $n \in \mathbb{N}$ and such that $\displaystyle{\lim_{n \to \infty} \| xy_n \| = 1}$ and $\displaystyle{\lim_{n \to \infty} \|y_nx \| = 1}$. Then:

\begin{align} \quad \inf \{ \| xy \| : \| y \| = 1 \} \leq \inf \{ \| xy_n \| : n \in \mathbb{N} \} = 0 \end{align}

- And:

\begin{align} \quad \inf \{ \| yx \| : \| y \| = 1 \} \leq \inf \{ \| y_nx \| : n \in \mathbb{N} \} = 0 \end{align}

- Thus $x$ is both a left topological divisor of zero and a right topological divisor of zero. Thus $x$ is a topological divisor of zero. $\blacksquare$