Boundary Points of Inv(X) are Topo. Divisors of Zero in a Banach Algebra

# Boundary Points of Inv(X) are Topological Divisors of Zero in a Banach Algebra

Recall from the Topological Divisors of Zero in a Normed Algebra page that if $X$ is a normed algebra then a point $x \in X$ is said to be a topological divisor of zero if:

(1)
\begin{align} \quad \inf \{ \| xy \| + \| yx \| : \| y \| = 1 \} = 0 \end{align}

Or equivalently, there exists a sequence $(y_n)$ with $\| y_n \| = 1$ for all $n \in \mathbb{N}$ such that both $\displaystyle{\lim_{n \to \infty} \| xy_n \| = 0}$ and $\displaystyle{\lim_{n \to \infty} \| y_nx\| = 0}$.

We will now prove that every point in the boundary of the set of invertible elements of $X$ will be a topological divisor of zero.

 Theorem 1: Let $X$ be a Banach algebra with unit. If $x \in \partial (\mathrm{Inv}(X))$ then $x$ is a topological divisor of zero.
• Proof: Let $x \in \partial (\mathrm{Inv}(X))$. Then there exists a sequence $(x_n)$ in $\mathrm{Inv}(X)$ that converges to $x$.
• Consider the numerical sequence $( \| x_n^{-1} \|)$. This sequence is unbounded. To prove this, suppose instead that there exists an $M > 0$ such that for every $n \in \mathbb{N}$:
(2)
\begin{align} \quad \| x_n^{-1} \| \leq M \end{align}
• Let $\epsilon > 0$ be given. Since $(x_n)$ converges to $x$ (and is hence Cauchy), for $\frac{\epsilon}{2M} > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\displaystyle{\| x_m - x_n \| < \frac{\epsilon}{M}}$. Thus if $m, n \geq N$ we also have that:
(3)
\begin{align} \quad \| x_m^{-1} - x_n^{-1} \| = \| x_m^{-1}(x_n - x_m)x_n^{-1} \| \leq \| x_m^{-1} \| \| x_n - x_m \| \| x_n^{-1} \| \leq 2M \| x_n - x_m \| < 2M \cdot \frac{\epsilon}{2M} = \epsilon \end{align}
• Thus $(x_n^{-1})$ is a Cauchy sequence. Since $X$ is a Banach algebra, $(x_n^{-1})$ converges to some $y \in X$. But observe that by continuity of multiplication that:
(4)
\begin{align} \quad xy = \left [ \lim_{n \to \infty} x_n \right ] \left [ \lim_{n \to \infty} x_n^{-1} \right ] = \lim_{n \to \infty} x_nx_n^{-1} = 1 \end{align}
• And also:
(5)
\begin{align} \quad yx = \left [ \lim_{n \to \infty} x_n^{-1} \right ] \left [ \lim_{n \to \infty} x_n \right ] = \lim_{n \to \infty} x_n^{-1}x_n = 1 \end{align}
• Thus $xy = yx = 1$. This shows that $x \in \mathrm{Inv}(X)$. But we know that $\mathrm{Inv}(X)$ is an open set from The Set of Invertible Elements Inv(X) is an Open Subset of X page, and an open set never contains any of its boundary points. So we have arrived at a contradiction. Thus the assumption that $(\| x_n^{-1} \|)$ is bounded is false. So $(\| x_n^{-1} \|)$ is unbounded.
• Since $(\| x_n^{-1} \|)$ is unbounded, for each $m \in \mathbb{N}$ there exists an $N_m \in \mathbb{N}$ such that if $n \geq N_m$ then:
(6)
\begin{align} \quad \| x_n^{-1} \| \geq m \end{align}
• Let $(y_n^{-1})$ be a subsequence of $(x_n^{-1})$ with the property that for every $n \in \mathbb{N}$, $\| y_n^{-1} \| \geq n$. For each $n \in \mathbb{N}$ let:
(7)
\begin{align} \quad z_n = \frac{y_n^{-1}}{\| y_n^{-1} \|} \end{align}
• Then $\| z_n \| = 1$ for all $n \in \mathbb{N}$, and furthermore we have that:
(8)
\begin{align} \quad \lim_{n \to \infty} \| xz_n \| = \lim_{n \to \infty} \| (x - y_n)z_n + y_nz_n \| \leq \lim_{n \to \infty} \| x - y_n \| \|z_n \| + \lim_{n \to \infty} \frac{\| 1 \|}{\| y_n^{-1} \|} = 0 + 0 = 0 \end{align}
• (Where $\lim_{n \to \infty} \| x - y_n \| \| z_n \| = 0$] since [[$(y_n)$ converges to $x$ and $\| z_n \| = 1$ for all $n \in \mathbb{N}$). Also:
(9)
\begin{align} \quad \lim_{n \to \infty} \| z_nx \| = \lim_{n \to \infty} \| z_n(x - y_n) + z_ny_n \| \leq \lim_{n \to \infty} \| z_n \| \| x - y_n \| + \lim_{n \to \infty} \frac{\| 1 \|}{\| y_n^{-1} \|}= 0 + 0 = 0 \end{align}
• So for the point $x \in \partial (\mathrm{Inv}(X))$ there exists a sequence $(z_n)$ with $\| z_n \| = 1$ for all $n \in \mathbb{N}$ such that both $\displaystyle{\lim_{n \to \infty} \| xz_n \| = 0}$ and $\displaystyle{\lim_{n \to \infty} \| z_nx \| = 0}$, and thus, $x$ is a topological divisor of zero. $\blacksquare$