The Dual Banach Right Module of a Normed Left A-Module M

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)

\begin{align} \quad (fa)(m) = f(am) \end{align}

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)

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

And for all $m \in M$ and $\alpha \in \mathbf{F}$:

(3)

\begin{align} \quad (fa)(\alpha m) = f(a \alpha m) = f(\alpha am) = \alpha f(am) = \alpha (fa)(m) \end{align}

Lastly, each $fa$ is bounded since for all $m \in M$:

(4)

\begin{align} \quad \| (fa)(m) \| = \| f(am) \| \leq \| f \| \| am \| \leq \| f \| \| a \| \| m \| \end{align}

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:

(5)

\begin{align} \quad ([f_1 + f_2]a)(m) = (f_1a + f_2a)(m) = f_1(am) + f_2(am) = (f_1a)(m) + (f_2a)(m) \quad m \in M \end{align}

• 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:

(6)

\begin{align} \quad (f[a_1 + a_2])(m) = (fa_1 + fa_2)(m) = f(a_1m) + f(a_2m) = (fa_1)(m) + (fa_2)(m) \quad m \in M \end{align}

• 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:

(7)

\begin{align} \quad ((fa_1)a_2)(m) = (fa_1)(a_2m) = f(a_1a_2m) = (fa_1a_2)(m) = (f(a_1a_2))(m) \quad m \in M \end{align}

• 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:

(8)

\begin{align} \quad \| (fa) \|_{\mathrm{op}} = \inf \{ M : |(fa)(m)| \leq M \| m \|_M, \: m \in M \} &= \inf \{ M : |f(am)| \leq M \| m \|_M, \: m \in M \} \end{align}

• 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:

(9)

\begin{align} \quad \| (fa) \|_{\mathrm{op}} \leq \| f \|_{\mathrm{op}} \| a \|_{\mathfrak{A}} \end{align}

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.