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

(5)While:

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