Bound. Pts. of Inv(A) are Topo. Div. of Zero in a Banach Algebra w/ Unit

Boundary Points of Inv(A) are Topological Divisors of Zero in a Banach Algebra with Unit

Recall from the Topological Divisors of Zero in a Normed Algebra page that if $\mathfrak{A}$ is a normed algebra then a point $x \in \mathfrak{A}$ 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 $\mathfrak{A}$ will be a topological divisor of zero.

Theorem 1: Let $\mathfrak{A}$ be a Banach algebra with unit. If $a \in \partial (\mathrm{Inv}(\mathfrak{A}))$ then $a$ is a topological divisor of zero.

• Proof: Let $a \in \partial (\mathrm{Inv}(\mathfrak{A}))$. Then there exists a sequence $(a_n)$ in $\mathrm{Inv}(\mathfrak{A})$ that converges to $a$.

• Consider the numerical sequence $( \| a_n^{-1} \|)$. 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 \| a_n^{-1} \| \leq M \end{align}

• Let $\epsilon > 0$ be given. Since $(a_n)$ converges to $a$ (and is hence Cauchy), for $\frac{\epsilon}{2M} > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\displaystyle{\| a_m - a_n \| < \frac{\epsilon}{M}}$. Thus if $m, n \geq N$ we also have that:

(3)

\begin{align} \quad \| a_m^{-1} - a_n^{-1} \| = \| a_m^{-1}(a_n - a_m)a_n^{-1} \| \leq \| a_m^{-1} \| \| a_n - a_m \| \| a_n^{-1} \| \leq 2M \| a_n - a_m \| < 2M \cdot \frac{\epsilon}{2M} = \epsilon \end{align}

• Thus $(a_n^{-1})$ is a Cauchy sequence. Since $\mathfrak{A}$ is a Banach algebra, $(a_n^{-1})$ converges to some $y \in \mathfrak{A}$. But observe that by continuity of multiplication that:

(4)

\begin{align} \quad ay = \left [ \lim_{n \to \infty} a_n \right ] \left [ \lim_{n \to \infty} a_n^{-1} \right ] = \lim_{n \to \infty} a_na_n^{-1} = 1 \end{align}

• And also:

(5)

\begin{align} \quad ya = \left [ \lim_{n \to \infty} a_n^{-1} \right ] \left [ \lim_{n \to \infty} a_n \right ] = \lim_{n \to \infty} a_n^{-1}a_n = 1 \end{align}

• Thus $ay = ya = 1$. This shows that $a \in \mathrm{Inv}(\mathfrak{A})$. But we know that $\mathrm{Inv}(\mathfrak{A})$ is an open set from The Set of Invertible Elements Inv(A) is an Open Subset of A page, and an open set never contains any of its boundary points. So we have arrived at a contradiction. Thus the assumption that $(\| a_n^{-1} \|)$ is bounded is false. So $(\| a_n^{-1} \|)$ is unbounded.

• Since $(\| a_n^{-1} \|)$ 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 \| a_n^{-1} \| \geq m \end{align}

• Let $(y_n^{-1})$ be a subsequence of $(a_n^{-1})$ with the property that for every $n \in \mathbb{N}$, $\| a_n^{-1} \| \geq n$. 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} \| az_n \| = \lim_{n \to \infty} \| (a - y_n)z_n + y_nz_n \| \leq \lim_{n \to \infty} \| a - y_n \| \|z_n \| + \lim_{n \to \infty} \frac{\| 1 \|}{\| y_n^{-1} \|} = 0 + 0 = 0 \end{align}

• (Where $\lim_{n \to \infty} \| a - y_n \| \| z_n \| = 0 $] since [[$ (y_n)$ converges to $a$ and $\| z_n \| = 1$ for all $n \in \mathbb{N}$). Also:

(9)

\begin{align} \quad \lim_{n \to \infty} \| z_na \| = \lim_{n \to \infty} \| z_n(a - y_n) + z_ny_n \| \leq \lim_{n \to \infty} \| z_n \| \| a - y_n \| + \lim_{n \to \infty} \frac{\| 1 \|}{\| y_n^{-1} \|}= 0 + 0 = 0 \end{align}

• So for the point $a \in \partial (\mathrm{Inv}(\mathfrak{A}))$ there exists a sequence $(z_n)$ with $\| z_n \| = 1$ for all $n \in \mathbb{N}$ such that both $\displaystyle{\lim_{n \to \infty} \| az_n \| = 0}$ and $\displaystyle{\lim_{n \to \infty} \| z_na \| = 0}$, and thus, $a$ is a topological divisor of zero. $\blacksquare$