Examples of Modules - c0(M) and l1(M) are Banach Left X-Modules

# Examples of Modules - For Normed Algebras X and Banach Left X-Module M, c0(M) and l1(M) are Banach Left X-Modules

## c_0 Banach Left X-Modules

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

(1)
\begin{align} \quad \| xm \| \leq K \| x \| \| m \| \end{align}

For all $x \in X$ 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 $c_0(M_{\alpha}, \alpha \in \Lambda)$ is a normed linear space.

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

(5)
\begin{align} \quad (xf)(\alpha) = xf(\alpha), \quad \alpha \in \Lambda \end{align}

Showing that for every $x \in X$ and for every $f \in c_0(M_{\alpha}, \alpha \in \Lambda)$ that $xf \in f \in c_0(M_{\alpha}, \alpha \in \Lambda)$:

We need to show that for every $x \in X$ and for every $f \in c_0(M_{\alpha}, \alpha \in \Lambda)$ that $xf \in c_0(M_{\alpha}, \alpha \in \Lambda)$.

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

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

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

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

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

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

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

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

Showing that $\| xf \| \leq K \| x \| \| f \|$:

Lastly, for all $f \in c_0(M_{\alpha}, \alpha \in \Lambda)$ and for all $x \in X$:

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

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

### c_0(M) as a Banach Left X-Module

Let $X$ be a Banach algebra and let $M$ be a Banach left $X$-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 $X$-module.

## l^1 Banach Left X-Modules

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

(10)
\begin{align} \quad \| xm \| \leq K \| x \| \| m \| \end{align}

For all $x \in X$ and for all $m \in M_{\alpha}$, $\alpha \in \Lambda$. Let:

(11)
\begin{align} \quad l^1(M_{\alpha} : \alpha \in \Lambda) = \left \{ f \in \prod_{\alpha \in \Lambda} M_{\alpha} : \sum_{\alpha \in \Lambda} \| f(\alpha) \| < \infty \right \} \end{align}

We equip this space with the operations of pointwise function addition and scalar multiplication to make it a linear space.

We define a norm on $l^1(M_{\alpha} : \alpha \in \Lambda)$ to make it a normed linear space. This norm is defined for all $f \in l^1(M_{\alpha} : \alpha \in \Lambda)$ by:

(12)
\begin{align} \quad \| f \| = \sum_{\alpha \in \Lambda} \| f(\alpha) \| \end{align}

$l^1(M_{\alpha} : \alpha \in \Lambda)$ becomes a Banach space with this norm. In fact, this space becomes a Banach left $X$-module with the module multiplication defined for all $f \in l^1(M_{\alpha} : \alpha \in \Lambda)$ and all $x \in X$ by:

(13)
\begin{align} \quad (xf)(\alpha) = xf(\alpha) \end{align}

(Again observe that $f(\alpha) \in M_{\alpha}$. Since $M_{\alpha}$ is a Banach left $X$-module we have that $xf(\alpha)$ is defined and furthermore that $xf(\alpha) \in M_{\alpha}$).

Showing that for all $x \in X$ and for all $f \in l^1(M_{\alpha} : \alpha \in \Lambda)$ that $xf \in l^1(M_{\alpha} : \alpha \in \Lambda)$:

We first need to show that for all $f \in l^1(M_{\alpha} : \alpha \in \Lambda)$ and for all $x \in X$ we have that $xf \in l^1(M_{\alpha} : \alpha \in \Lambda)$, that is, we want to show that:

But this is simple since:

(14)
\begin{align} \quad \sum_{\alpha \in \Lambda} \| (xf)(\alpha) \| = \sum_{\alpha \in \Lambda} \| xf(\alpha) \| \leq \sum_{\alpha \in \Lambda} K \| x \| \| f(\alpha) \| = K \| x \| \underbrace{\sum_{\alpha \in \Lambda} \| f(\alpha) \|}_{< \infty} < \infty \end{align}

So $xf \in l^1(M_{\alpha} : \alpha \in \Lambda)$.

Showing that $\| xf \| \leq K \| x \| \| f \|$:

Again we have that:

(15)
\begin{align} \quad \| xf \| = \sum_{\alpha \in \Lambda} \| (xf)(\alpha) \| \leq \sum_{\alpha \in \Lambda} \| xf(\alpha) \| \leq \sum_{\alpha \in \Lambda} K \| x \| \| f(\alpha) \| = K \| x \| \sum_{\alpha \in \Lambda} \| f(\alpha) \| = K \| x \| \| f \| \end{align}

So $l^1(M_{\alpha} : \alpha \in \Lambda)$ is a Banach left $X$-module.

### l^1(M) as a Banach Left X-Module

Again by taking $\Lambda = \mathbb{N}$ and $M_n = M$ for all $n \in \mathbb{N}$ we obtain $l^1(M)$ as a Banach left $X$-module.