Examples of Modules - The Dual Banach Right Module of a Normed Left A-Module M
Let $\mathfrak{A}$ be a normed algebra and let $M$ be a normed left $\mathfrak{A}$-module. Let $M^*$ be the dual space of $M$ (the space of all bounded linear functionals on $M$). Then $M^*$ is a Banach right $\mathfrak{A}$-module called the Dual Banach Right Module of $M$, with the following module multiplication from $\mathfrak{A} \times M^* \to M^*$ as follows.
For each $a \in \mathfrak{A}$ and each $f \in M^*$ we define $(fa)$ to be the functional defined for all $m \in M$ by:
(1)Note that $(fa)$ is well-defined. Since $M$ is a left $\mathfrak{A}$-module we have that the multiplication $am$ is defined and contained in $M$ (which is significant since $f \in M^*$).
It is first important to check that each $fa$ is also in $M^*$.
First $fa$ is linear since for all $m_1, m_2 \in M$ we have that:
(2)And for all $m \in M$ and $\alpha \in \mathbf{F}$:
(3)Lastly, each $fa$ is bounded since for all $m \in M$:
(4)So $(fa)$ is bounded with $\| fa \| \leq \| f \| \| a \|$. So $(fa) \in M^*$.
We now check that $M^*$ is a Banach right $\mathfrak{A}$-module:
- 1. For each fixed $a \in \mathfrak{A}$, we have that for all $f_1, f_2 \in M^*$ that:
- And thus for each $a \in \mathfrak{A}$, multiplication $(a, f) \to (fa)$ is linear and so $RM1$ is satisfied.
- 2. Similarly, for each fixed $f \in M^*$ we have that for all $a_1, a_2 \in \mathfrak{A}$ that:
- And thus for each $f \in M^*$, multiplication $(a, f) \to (fa)$ is linear and so $RM2$ is satisfied.
- 3. Lastly, for all $a_1, a_2 \in \mathfrak{A}$ and all $f \in M^*$ we have that:
- So $RM3$ is satisfied and so $M^*$ is an right $\mathfrak{A}$-module.
- 4. For all $a \in \mathfrak{A}$ and all $f \in M^*$ we have that:
- Note that since $f$ is a bounded linear functional we have that $|f(am)| \leq \| f \| \| am \|_M \leq \| f \|_{\mathrm{op}} \| a \|_{\mathfrak{A}} \| m \|_M$, and thus:
So $M^*$ is a Banach right $\mathfrak{A}$-module.
It can similarly be shown that if $\mathfrak{A}$ is a normed algebra and $M$ is a normed right $\mathfrak{A}$-module then $M^*$ is a Banach left $\mathfrak{A}$-module.