Examples of Modules - For L. X-Mod. M, M* is a Banach R. X-Mod.

Examples of Modules - For Normed Algebras X and Left X-Modules M, M* is a Banach Right X-Module

Let $X$ be a normed algebra and let $M$ be a normed left $X$-module. Let $M^*$ be the dual space of $M$ (the space of all bounded linear functionals on $M$). Then $M^*$ is a right $X$-module with the following module multiplication from $X \times M^* \to M^*$ as follows.

For each $x \in X$ and each $f \in M^*$ we define $(fx)$ to be the functional defined for all $m \in M$ by:

(1)
\begin{align} \quad (fx)(m) = f(xm) \end{align}

Note that $(fx)$ is well-defined. Since $M$ is a left $X$-module we have that the multiplication $xm$ is defined and contained in $M$ (which is significant since $f \in M^*$).

Showing that for every $f \in M^*$ and every $x \in X$ that $(fx) \in M^*$:

For linearity, let $m_1, m_2 \in M$. Then:

(2)
\begin{align} \quad (fx)(m_1 + m_2) = f(x[m_1 + m_2]) = f(xm_1 + xm_2) = f(xm_1) + f(xm_2) = (fx)(m_1) + (fx)(m_2) \end{align}

Let $m \in M$ and $a \in \mathbf{F}$. Then:

(3)
\begin{align} \quad (fx)(am) = f(xam) = f(axm) = af(xm) = a(fx)(m) \end{align}

And for boundedness we have that for every $m \in M$:

(4)
\begin{align} \quad \| (fx)(m) \| = \| f(xm) \| \leq \| f \| \| xm \| \leq \| f \| \| x \| \| m \| \end{align}

So $(fx)$ is bounded with $\| fx \| \leq \| f \| \| x \|$. So $(fx) \in M^*$.

Checking the axioms $LM1$, $LM2$, and $LM3$:

For each fixed $x \in X$, we have that for all $f_1, f_2 \in M^*$ that:

(5)
\begin{align} \quad ([f_1 + f_2]x)(m) = (f_1x + f_2x)(m) = f_1(xm) + f_2(xm) = (f_1x)(m) + (f_2x)(m) \quad m \in M \end{align}

And thus for each $x \in X$, multiplication $(x, f) \to (fx)$ is linear and so $LM1$ is satisfied.

Similarly, for each fixed $f \in M^*$ we have that for all $x_1, x_2 \in X$ that:

(6)
\begin{align} \quad (f[x_1 + x_2])(m) = (fx_1 + fx_2)(m) = f(x_1m) + f(x_2m) = (fx_1)(m) + (fx_2)(m) \quad m \in M \end{align}

And thus for each $f \in M^*$, multiplication $(x, f) \to (fx)$ is linear and so $LM2$ is satisfied.

Lastly, for all $x_1, x_2 \in X$ and all $f \in M^*$ we have that:

(7)
\begin{align} \quad ((fx_1)x_2)(m) = (fx_1)(x_2m) = f(x_1x_2m) = (fx_1x_2)(m) = (f(x_1x_2))(m) \quad m \in M \end{align}

So $LM3$ is satisfied and so $M^*$ is an right $X$-module.

Showing that there exists a $K > 0$ such that $\| (fx) \| \leq K \| f \| \| x \|$ for all $f \in M^*$ and all $x \in X$:

It is furthermore a normed right $X$-module since for all $x \in X$ and all $f \in M^*$ we have that:

(8)
\begin{align} \quad \| (fx) \|_{\mathrm{op}} = \inf \{ M : |(fx)(m)| \leq M \| m \|_M, \: m \in M \} &= \inf \{ M : |f(xm)| \leq M \| m \|_M, \: m \in M \} \end{align}

Note that since $f$ is a bounded linear functional we have that $|f(xm)| \leq \| f \| \| xm \|_M \leq \| f \|_{\mathrm{op}} \| x \|_X \| m \|_M$, and thus:

(9)
\begin{align} \quad \| (fx) \|_{\mathrm{op}} \leq \| f \|_{\mathrm{op}} \| x \|_X \end{align}

So $M^*$ is a normed right $X$-module and furthermore, since $M^* = \mathcal B(M, \mathbf{F})$ is always a Banach space we have that $M^*$ is a Banach right $X$-module.

It can similarly be shown that if $X$ is a normed algebra and $M$ is a normed right $X$-module then $M^*$ is a Banach left $X$-module.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License