Approximate Identities in A for X when X is a Banach Left A-Module 4

Recall from the Approximate Identities in A for X when X is a Banach Left A-Module 3 page the following theorem: Let $\mathfrak{A}$ be a Banach algebra and let $X$ be a Banach left $\mathfrak{A}$-module. If $\mathfrak{A}$ has a bounded left approximate identity for $X$ and $z \in X$, $\delta > 0$ then there exists $a \in \mathfrak{A}$ and $y \in X$ such that $z = ay$ and $\| z - y \| \leq \delta$.

We now prove some important corollaries to this result.

Corollary 1: Let $\mathfrak{A}$ be a Banach algebra. If $\mathfrak{A}$ has a bounded left approximate identity and $z \in \mathfrak{A}$ and $\delta > 0$ then there exists $x, y \in \mathfrak{A}$ where $y$ is contained in the least closed left ideal of $\mathfrak{A}$ containing $z$, $z = xy$, and $\| x - y \| \leq \delta$.

• Proof: Let $X$ be the least closed left ideal containing $z$. Then $X\mathfrak{A} \subseteq X \subseteq \mathfrak{A}$. So $X$ can be regarded as a Banach left $\mathfrak{A}$-module.

• By Theorem 1, for the point $z$ there exists $x \in \mathfrak{A}$, $y \in X$ such that $z = xy$ and $\| x - y \| \leq \delta$. And of course, $y \in X$ simply says $y$ is contained in the least closed left ideal of $\mathfrak{A}$. $\blacksquare$

Corollary 2: Let $\mathfrak{A}$ be a Banach algebra. If $\mathfrak{A}$ has a bounded left approximate identity $\{ e(\lambda) \}_{\lambda \in \Lambda}$ and $(z_n) \subset \mathfrak{A}$ is such that $\displaystyle{\lim_{n \to \infty} z_n = 0}$ then there exists $a \in \mathfrak{A}$ and $(y_n) \subset \mathfrak{A}$ such that $z_n = ay_n$ for every $n \in \mathbb{N}$ and $\displaystyle{\lim_{n \to \infty} y_n = 0}$.

• Proof: Let

(1)

\begin{align} \quad X = \left \{ (a_n) \subset \mathfrak{A} : \lim_{n \to \infty} a_n = 0 \right \} \end{align}

• Equip $X$ with the operations of sequence addition, scalar multiplication, and define a norm on $X$ for all $(a_n) \in X$ by:

(2)

\begin{align} \quad \| (a_n) \|_X = \sup_{n \in \mathbb{N}} \{ \| a_n \|_{\mathfrak{A}} \} \end{align}

• (Note that the supremum exists since $(\| (a_n) \|_{\mathfrak{A}})$ is a bounded numerical sequence.)

• Then $X$ can be seen as a Banach left $\mathfrak{A}$-module with module multiplication defined for all $a \in {\mathfrak{A}}$ and all $(a_n) \in X$ by:

(3)

\begin{align} \quad a(a_n) = (aa_n) \end{align}

• We aim to show that $e(\lambda)(a_n) - (a_n) \to 0$ for all $(a_n) \in X$. We prove this below.

• First since $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is bounded there exists a $K > 0$ such that $\| e(\lambda) \|_{\mathfrak{A}} \leq K$.

• Let $\epsilon > 0$ be given and let $(a_n) \subseteq X$. Then $\lim_{n \to \infty} a_n = 0$, that is, $\lim_{n \to \infty} \| a_n \|_{\mathfrak{A}} = 0$. So for $\frac{\epsilon}{K + 1} > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\| a_n \|_{\mathfrak{A}} < \frac{\epsilon}{K+1}$. So if $n \geq N$ then:

(4)

\begin{align} \quad \| e(\lambda)(a_n) - (a_n) \|_X &= \| (e(\lambda)a_n - a_n) \|_X \\ &= \sup_{n \in \mathbb{N}} \left \{ \| e(\lambda)a_n - a_n \|_{\mathfrak{A}} \right \} \\ & \leq \sup_{n \in \mathbb{N}} \left \{ (\| e(\lambda) \|_{\mathfrak{A}} + 1) \| a_n \|_{\mathfrak{A}} \right \} \\ & \leq [\| e(\lambda) \|_{\mathfrak{A}} + 1] \sup_{n \in \mathbb{N}} \left \{ \| a_n \|_{\mathfrak{A}} \right \} \\ & \leq [K + 1] \| (a_n) \|_X \\ & < [K + 1] \frac{\epsilon}{K + 1} \\ & < \epsilon \end{align}

• So $e(\lambda)(a_n) - (a_n) \to 0$ for all $(a_n) \in X$, that is, $\{ e(\lambda) \}_{\lambda \in \Lambda}$ is a bounded left approximate identity in $\mathfrak{A}$ for $X$.

• So by Theorem 1, for the sequence $(z_n) \in X$ there exists an $a \in \mathfrak{A}$ and a sequence $(y_n) \in X$ such that $(z_n) = a(y_n)$, that is, $z_n = ay_n$ for each $n \in \mathbb{N}$. Furthermore, since $(y_n) \in X$ by definition of $X$ we have that $\displaystyle{\lim_{n \to \infty} y_n = 0}$. $\blacksquare$