For Banach Left A-Modules M C0(M) is a Banach Left A-Module

# Examples of Modules - For Banach Left A-Modules M C0(M) is a Banach Left A-Module

Let $\mathfrak{A}$ be Banach algebra and let $\{ M_{\alpha} : \alpha \in \Lambda \}$ be a collection of Banach left $\mathfrak{A}$-modules such that there exists a $K > 0$ such that:

(1)
\begin{align} \quad \| am \| \leq K \| a \| \| m \| \end{align}

for all $a \in \mathfrak{A}$ and for all $m \in M_{\alpha}$, $\alpha \in \Lambda$.

Let $c_0 ( M_{\alpha} : \alpha \in \Lambda)$ be the set of all functions $f$ on $\prod_{\alpha \in \Lambda} M_{\alpha}$ such that for all $\epsilon > 0$ we have that:

(2)
\begin{align} \quad \left | \{ \alpha \in \Lambda : \| f(\alpha) \| > \epsilon \} \right | < \infty \quad (*) \end{align}

On the set $c_0(M_{\alpha}, \alpha \in \Lambda)$ define the operations of pointwise function addition and scalar multiplication for all $\alpha \in \Lambda$ by:

(3)
\begin{align} \quad [f + g](\alpha) &= f(\alpha) + g(\alpha) \\ \quad f(t \alpha) &= t f(\alpha) \end{align}

Then $c_0(M_{\alpha}, \alpha \in \Lambda)$ is a linear space. Moreover, we define a norm on $c_0(M_{\alpha}, \alpha \in \Lambda)$ for all $f \in c_0(M_{\alpha}, \alpha \in \Lambda)$ by:

(4)
\begin{align} \quad \| f \| = \sup \{ \| f(\alpha) \| : \alpha \in \Lambda \} \end{align}

This norm is guaranteed to be well-defined by $(*)$, and makes $c_0(M_{\alpha}, \alpha \in \Lambda)$ a normed linear space.

Now $c_0(M_{\alpha}, \alpha \in \Lambda)$ will become a Banach Left $\mathfrak{A}$-module with module multiplication $\mathfrak{A} \times c_0(M_{\alpha}, \alpha \in \Lambda) \to c_0(M_{\alpha}, \alpha \in \Lambda)$ defined for all $a \in \mathfrak{A}$ and all $f \in c_0(M_{\alpha}, \alpha \in \Lambda)$ by:

(5)

We need to show that for every $a \in \mathfrak{A}$ and for every $f \in c_0(M_{\alpha}, \alpha \in \Lambda)$ that $af \in c_0(M_{\alpha}, \alpha \in \Lambda)$.

First we show that $af \in \prod_{\alpha \in \Lambda} M_{\alpha}$. For each $\alpha \in \Lambda$ we have that $(af)(\alpha) = af(\alpha)$. Since $a \in \mathfrak{A}$ and $f(\alpha) \in M_{\alpha}$ we see that $af(\alpha)$ is just taking an element $f(\alpha)$ of $M_{\alpha}$ and multiplying it on the left by an element $a$ of $\mathfrak{A}$. Since $M_{\alpha}$ is a Banach left $\mathfrak{A}$-module, the multiplication $af(\alpha)$ is defined and $af(\alpha) \in M_{\alpha}$. So indeed, $af \in \prod_{\alpha \in \Lambda}$.

We now show that $af \in c_0(M_{\alpha}, \alpha \in \Lambda)$. Let $\epsilon > 0$ be given and consider the set:

(6)
\begin{align} \quad \{ \alpha \in \Lambda : \| (af)(\alpha) \| > \epsilon \} &= \{ \alpha \in \Lambda : \| af(\alpha) \| > \epsilon \} \end{align}

If $a = 0$ then observe that $\| (af)(\alpha) \| = \| af(\alpha) \| \leq K \| a \| \| f(\alpha) \| = 0$. So clearly:

(7)
\begin{align} \quad \left | \{ \alpha \in \Lambda : \| (af)(\alpha) \| > \epsilon \} \right | = 0 < \infty \end{align}

If $a \neq 0$ then the case is a bit different. Note that if $\alpha \in \Lambda$ is such that $\epsilon < \| af(\alpha) \|$ then since $\| af(\alpha) \| \leq K \| a \| \| f(\alpha) \|$ we have that $\epsilon < K \| a \| \| f(\alpha) \|$ too. So:

(8)
\begin{align} \quad \{ \alpha \in \Lambda : \| (af)(\alpha) \| > \epsilon \} \subseteq \{ \alpha \in \Lambda : K \| a \| \| f(\alpha) \| > \epsilon \} = \left \{ \alpha \in \Lambda : \| f(\alpha) \| > \frac{\epsilon}{\| a \| K} \right \} \end{align}

The far righthand set is finite since $f \in c_0(M_{\alpha}, \alpha \in \Lambda)$. Therefore $\{ \alpha \in \Lambda : \| (af)(\alpha) \| > \epsilon \}$ is a finite set for every $\epsilon > 0$ and so $(af) \in c_0(M_{\alpha}, \alpha \in \Lambda)$.

Lastly, for all $f \in c_0(M_{\alpha}, \alpha \in \Lambda)$ and for all $a \in \mathfrak{A}$:

(9)
\begin{align} \quad \| af \| &= \sup_{\alpha \in \Lambda} \{ \| (af)(\alpha) \| \} = \sup_{\alpha \in \Lambda} \{ \| af(\alpha) \| \} \leq \sup_{\alpha \in \Lambda} \{ K \| a \| \| f(\alpha) \| \} = K \| a \| \sup_{\alpha \in \Lambda} \{ \| f(\alpha) \| \} = \leq K \| a \| \| f \| \end{align}

So indeed $c_0(M_{\alpha}, \alpha \in \Lambda)$ is a Banach left $\mathfrak{A}$-module.

Let $\mathfrak{A}$ be a Banach algebra and let $M$ be a Banach left $\mathfrak{A}$-module. Let $\Lambda = \mathbb{N}$. For each $n \in \mathbb{N}$ let $M_n = M$. We denote $c_0(M) = c_0(M_n : n \in \mathbb{N})$, which consists of all infinite sequences whose terms are elements of $M$.

As with above, $c_0(M)$ is a Banach left $\mathfrak{A}$-module.