For Normed Left A-Modules M, M is a Normed Left (A + F)-Module

Examples of Modules - For Normed Left A-Modules M, M is a Left (A + F)-Module

Let $\mathfrak{A}$ be a normed algebra and let $M$ be a normed left $\mathfrak{A}$-module. Recall that the unitization $\mathfrak{A} + \mathbf{F}$ is normed algebra with unit $(0, 1)$ with norm defined on $\mathfrak{A} + \mathbf{F}$ for all $(a, \alpha) \in \mathfrak{A} + \mathbf{F}$ by:

(1)

\begin{align} \quad \| (a, \alpha) \| = \| a \| + |\alpha| \end{align}

As we will see, $M$ will become an left ($\mathfrak{A} + \mathbf{F}$)-module with module multiplication defined for all $m \in M$ and for all $(a, \alpha) \in \mathfrak{A} + \mathbf{F}$ by:

(2)

\begin{align} \quad (a, \alpha)m = am + \alpha m \end{align}

Note that indeed this module multiplication takes elements of $(\mathfrak{A} + \mathbf{F}) \times M$ and outputs elements of $M$. To see this, observe that since $M$ is a left $\mathfrak{A}$-module we have that $am \in M$. Furthermore, since $M$ is by definition also a linear space we have that $\alpha m \in M$. Thus $am + \alpha m \in M$.

• 1. For each fixed $(a, \alpha) \in \mathfrak{A} + \mathbf{F}$ we have that $((a, \alpha), m) \to (a, \alpha)m$ is linear since for all $m_1, m_2 \in M$ we have that:

(3)

\begin{align} \quad \quad (a, \alpha)[m_1 + m_2] = a[m_1 + m_2] + \alpha[m_1 + m_2] = am_1 + am_2 + \alpha m_1 + \alpha m_2 = (am_1 + \alpha m_1) + (am_2 + \alpha m_2) = (a, \alpha)m_1 + (a, \alpha)m_2 \end{align}

• 2. And for each fixed $m \in M$ we have that $((a, \alpha), m) \to (a, \alpha)m$ is linear since for all $(a, \alpha), (b, \beta) \in \mathfrak{A} + \mathbf{F}$ we have that:

(4)

\begin{align} \quad \quad [(a, \alpha) + (b, \beta)]m = (a + b, \alpha + \beta)m = (a + b)m + (\alpha + \beta)m = am + bm + \alpha m + \beta m = (am + \alpha m) + (bm + \beta m) = (a, \alpha)m + (b, \beta)m \end{align}

• 3. We have that for all $(a, \alpha), (b, \beta) \in \mathfrak{A} + \mathbf{F}$ and all $m \in M$:

(5)

\begin{align} \quad (a, \alpha)[(b, \beta)m] = (a, \alpha)[bm + \beta m] &= a[bm + \beta m] + \alpha [bm + \beta m] \\ &= abm + \beta am + \alpha bm + \alpha \beta m \end{align}

• While:

(6)

\begin{align} \quad [(a, \alpha)(b, \beta)]m = (ab + \alpha b + \beta a, \alpha \beta)m &= [ab + \alpha b + \beta a]m + \alpha \beta m \\ &= abm + \alpha bm + \beta am + \alpha \beta m \end{align}

• So $(a, \alpha)[(b, \beta)m] = [(a, \alpha)(b, \beta)]m$. Thus $M$ is a normed left ($\mathfrak{A} + \mathbf{F}$)-module.

• 4. Lastly, for all $(a, \alpha) \in \mathfrak{A} + \mathbf{F}$ and for all $m \in M$ we have that:

(7)

\begin{align} \quad \| (a, \alpha)m \|_M = \| am + \alpha m \|_M \leq \| am \|_M + \| \alpha m \|_M \leq \| a \|_{\mathfrak{A}} \| m \|_M + |\alpha| \| m \|_M = [\| a \|_{\mathfrak{A}} + |\alpha|] \| m \|_M = \| (a, \alpha) \|_{\mathfrak{A} + \mathbf{F}} \| m \|_M \end{align}

So we conclude that $M$ is a normed left $(\mathfrak{A} + \mathbf{F})$-module.

A similar argument shows that if $M$ is a normed right $\mathfrak{A}$-module then $M$ is also a normed right $(\mathfrak{A} + \mathbf{F})$-module.