Examples of Mods - For Norm. Left X-Mods. M, M is a Left (X + F)-Mod.

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

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

(1)
\begin{align} \quad \| (x, \alpha) \| = \| x \| + |\alpha| \end{align}

$M$ will become an left ($X + \mathbf{F}$)-module with module multiplication defined for all $m \in M$ and for all $(x, \alpha) \in X + \mathbf{F}$ by:

(2)
\begin{align} \quad (x, \alpha)m = xm + \alpha m \end{align}

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

For each fixed $(x, \alpha) \in X + \mathbf{F}$ we have that $((x, \alpha), m) \to (x, \alpha)m$ is linear since for all $m_1, m_2 \in M$ we have that:

(3)
\begin{align} \quad \quad (x, \alpha)[m_1 + m_2] = x[m_1 + m_2] + \alpha[m_1 + m_2] = xm_1 + xm_2 + \alpha m_1 + \alpha m_2 = (xm_1 + \alpha m_1) + (xm_2 + \alpha m_2) = (x, \alpha)m_1 + (x, \alpha)m_2 \end{align}

And for each fixed $m \in M$ we have that $((x, \alpha), m) \to (x, \alpha)m$ is linear since for all $(x, \alpha), (y, \beta) \in X + \mathbf{F}$ we have that:

(4)
\begin{align} \quad \quad [(x, \alpha) + (y, \beta)]m = (x + y, \alpha + \beta)m = (x + y)m + (\alpha + \beta)m = xm + ym + \alpha m + \beta m = (xm + \alpha m) + (ym + \beta m) = (x, \alpha)m + (y, \beta)m \end{align}

Lastly we have that for all $(x, \alpha), (y, \beta) \in X + \mathbf{F}$ and all $m \in M$:

(5)
\begin{align} \quad (x, \alpha)[(y, \beta)m] = (x, \alpha)[ym + \beta m] &= x[ym + \beta m] + \alpha [ym + \beta m] \\ &= xym + \beta xm + \alpha ym + \alpha \beta m \end{align}

While:

(6)
\begin{align} \quad [(x, \alpha)(y, \beta)]m = (xy + \alpha y + \beta x, \alpha \beta)m &= [xy + \alpha y + \beta x]m + \alpha \beta m \\ &= xym + \alpha ym + \beta xm + \alpha \beta m \end{align}

So $(x, \alpha)[(y, \beta)m] = [(x, \alpha)(y, \beta)]m$. Thus $M$ is a normed left ($X + \mathbf{F}$)-module.

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