Examples of Modules - For Norm. Alge. X, X* is a Banach L. X**-Bimod.

# Examples of Modules - For Normed Algebras X, X* is a Banach Left X**-Module

Let $X$ be a normed algebra. We will see that the second dual $X^{**}$ can be made into a Banach algebra with the operations of addition, scalar multiplication, and a special type of multiplication called the Arens product. The dual space $X^*$ can be regarded as a Banach left $X^{**}$-module with module multiplication defined for all $\varphi \in X^{**}$ and all $f \in X^*$ via $(\varphi, f) \to \varphi f$ where $\varphi f \in X^*$ is the bounded linear functional on $X$ defined for all $x \in X$ by $(\varphi f)(x) = \varphi(fx)$ where $fx \in X^*$ is the bounded linear functional on $X$ defined for all $y \in X$ by $(fx)(y) = f(xy)$.

For each $f \in X^*$ and $x \in X$ let $fx : X \to \mathbf{F}$ be defined for all $y \in X$ by:

(1)
\begin{align} \quad (fx)(y) = f(xy) \end{align}
 Proposition 1: Let $X$ be a normed algebra. If $f \in X^*$ and $x \in X$ then the map $fx : X \to \mathbf{F}$ is a bounded linear operator on $X$.
• Proof: There are two things to prove.
• 1. Showing that $(fx)$ is linear. Let $y, z \in X$ and let $\alpha \in \mathbf{F}$. Then by the definition of $fx$ and the linearity of $f$ we have that:
(2)
\begin{align} \quad (fx)(y + z) = f(x(y + z)) = f(xy +xz) = f(xy) + f(xz) = (fx)(y) + (fx)(z) \end{align}
(3)
\begin{align} \quad (fx)(\alpha y) = f(x(\alpha y)) = f(\alpha xy) = \alpha f(xy) = \alpha (fx)(y) \end{align}
• 2. Showing that $(fx)$ is bounded. For all $y \in X$ we have by the definition of $fx$ and the boundedness of $f$ that:
(4)
\begin{align} \quad | (fx)(y) | = | f(xy) | \leq \| f \| \| xy \| \leq \| f \| \| x \| \| y \| \end{align}
• So $(fx)$ is bounded and $\| fx \| \leq \| f \| \| x \|$. So $fx \in X^*$. $\blacksquare$

Now for $\varphi \in X^{**}$, $f \in X^*$, let $\varphi f : X \to \mathbf{F}$ be defined for all $x \in X$ by:

(5)
\begin{align} \quad (\varphi f)(x) = \varphi (fx) \end{align}
 Proposition 2: Let $X$ be a normed algebra. If $\varphi \in X^{**}$ and $f \in X^*$ then the map $\varphi f : X \to \mathbf{R}$ is a bounded linear operator
• Proof: There are two parts to this proof.
• 1. Showing that $(\varphi f)$ is linear: Let $x_1, x_2 \in X$. Then by the linearity of $\varphi$ and the fact that:
(6)
\begin{align} \quad (\varphi f)(x_1 + x_2) = \varphi(f[x_1 + x_2]) \overset{(*)} = \varphi (fx_1 + fx_2) \overset{(**)} = \varphi(fx_1) + \varphi(fx_2) \end{align}
• Now for all $x \in X$ and all $\alpha \in \mathbf{F}$ we have that:
(7)
\begin{align} \quad (\varphi f)(\alpha x) = \varphi(f[\alpha x]) = \varphi(\alpha fx) = \alpha \varphi (fx) \end{align}
• 2. Showing that $(\varphi f)$ is bounded. For all $x \in X$ we have that:
(8)
\begin{align} \quad \quad | (\varphi f)(x) | = | \varphi (fx) | \leq \| \varphi \| \| fx \| \overset{(\mathrm{Proposition \: 1})}\leq \| \varphi \| \| f \| \| x \| \end{align}
• So $\varphi f$ is bounded with $\| \varphi f \| \leq \| \varphi \| \| f \|$. So $\varphi f \in X^*$. $\blacksquare$

Let $X$ be a normed algebra. Given $\varphi, \psi \in X^{**}$, the Arens Product of $\varphi$ and $\psi$ denoted $(\varphi \psi) : X^* \to \mathbf{F}$ defined for all $f, g \in X^*$ by:

(9)
\begin{align} (\varphi \psi)(f) = \varphi( \psi f) \end{align}
 Proposition 3: Let $X$ be a normed algebra. If $\varphi, \psi \in X^{**}$ then the Arens product $(\varphi \psi)$ is a bounded linear operator on $X^*$.
• Proof: There are two parts to this proof.
• 1. Showing that $(\varphi \psi)$ is linear. Let $f, g \in X^*$ and let $\alpha \in \mathbf{F}$. Then:
(10)
\begin{align} \quad (\varphi \psi)(f + g) = \varphi(\psi[f + g]) = \varphi(\psi f + \psi g) = \varphi(\psi f) + \varphi(\psi g) = (\varphi \psi)(f) + (\varphi \psi)(g) \end{align}
(11)
\begin{align} \quad (\varphi \psi)(\alpha f) = \varphi(\psi \alpha f) = \varphi (\alpha \psi f) = \alpha \varphi (\psi f) = \alpha (\varphi \psi)(f) \end{align}
• 2: Showing that $\varphi \psi$ is bounded. For all $f \in X^*$ we have that:
(12)
\begin{align} \quad \| (\varphi \psi)(f) \| = \| \varphi (\psi f) \| \leq \| \varphi \| \| \psi f \| \overset{(\mathrm{Proposition \: 2})} \leq \| \varphi \| \| \psi \| \| f \| \end{align}
• So $\varphi \psi$ is bounded with $\| \varphi \psi \| \leq \| \varphi \| \| \psi \|$. So $\varphi \psi \in X^{**}$. $\blacksquare$

And finally:

 Proposition 4: Let $X$ be a normed algebra. Then $X^{**}$ with the operations of function addition, scalar multiplication, and the Arens product is a Banach algebra.

This is a tedious routine so we will omit it.

Now $X^*$ with become an Banach left $X^{**}$-module with module multiplication given for all $\varphi \in X^{**}$ and all $f \in X^*$ by $(\varphi f)(x) = \varphi(fx)$ (as defined earlier).

Showing that for all $\varphi \in X^{**}$ and for all $f \in X^*$ we have that $\varphi f \in X^*$:

As mentioned above, the multiplication $X^{**} \times X^* \to X^*$ defined above does multiply an element $\varphi$ of $X^{**}$ with an element $f$ of $X^*$ so that the product $\varphi f$ is in $X^*$ (by proposition 2).

Showing that there exists a $K > 0$ such that $\| \varphi f \| \leq K \| \varphi \| \| f \|$ for all $\varphi \in X^{**}$ and all $f \in X^*$:

We note that for every $x \in X$ that:

(13)
\begin{align} \quad | \varphi (fx) | \leq \| \varphi \| \| fx \| &= \| \varphi \| \inf \{ M : | (fx)(y) | \leq M \| y \|, \: \forall y \in X \} \\ &= \| \varphi \| \inf \{ M : |f(xy)| \leq M \| y \|, \forall y \in X \} \end{align}

Clearly $|f(xy)| \leq \| f \| \| xy \| \leq \| f \| \| x \| \| y \|$ and so:

(14)
\begin{align} \quad | \varphi (fx) | \leq \| \varphi \| \| f \| \| x \| \end{align}

Using the above we see that:

(15)
\begin{align} \quad \| \varphi f \| = \inf \{ M : | (\varphi f)(x) | \leq M \| x \|, \: \forall x \in X \} = \inf \{ M : | \varphi (fx) | \leq M \| x \|, \: \forall x \in X \} \leq \| \varphi \| \| f \| \end{align}

So with setting $K = 1$ we see that $\| \varphi f \| \leq K \| \varphi \| \| f \|$ for all $\varphi \in X^{**}$ and all $f \in X^*$.

So $X^*$ is a Banach left $X^{**}$-module.