For Banach Left A-Modules M, l1(M) is a Banach Left A-Module

# Examples of Modules - For Banach Left A-Modules M, l1(M) is a Banach Left A-Module

Let $\mathfrak{A}$ be a Banach algebra and again 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:

(2)
\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:

(3)
\begin{align} \quad \| f \|_1 = \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 $\mathfrak{A}$-module with the module multiplication defined for all $f \in l^1(M_{\alpha} : \alpha \in \Lambda)$ and all $a \in \mathfrak{A}$ by:

(4)
\begin{align} \quad (af)(\alpha) = af(\alpha) \end{align}

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

Showing that for all $a \in \mathfrak{A}$ and for all $f \in l^1(M_{\alpha} : \alpha \in \Lambda)$ that $af \in l^1(M_{\alpha} : \alpha \in \Lambda)$:

We first need to show that for all $f \in l^1(M_{\alpha} : \alpha \in \Lambda)$ and for all $a \in \mathfrak{A}$ we have that $af \in l^1(M_{\alpha} : \alpha \in \Lambda)$. We have that:

(5)
\begin{align} \quad \sum_{\alpha \in \Lambda} \| (af)(\alpha) \| = \sum_{\alpha \in \Lambda} \| af(\alpha) \| \leq \sum_{\alpha \in \Lambda} K \| a \| \| f(\alpha) \| = K \| a \| \underbrace{\sum_{\alpha \in \Lambda} \| f(\alpha) \|}_{< \infty} < \infty \end{align}

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

Lastly we have that:

(6)
\begin{align} \quad \| af \| = \sum_{\alpha \in \Lambda} \| (af)(\alpha) \| \leq \sum_{\alpha \in \Lambda} \| af(\alpha) \| \leq \sum_{\alpha \in \Lambda} K \| a \| \| f(\alpha) \| = K \| a \| \sum_{\alpha \in \Lambda} \| f(\alpha) \| = K \| a \| \| f \| \end{align}

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

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

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