For Normed Algebras A, A* is a Banach Left A**-Bimodule

# Examples of Modules - For Normed Algebras A, A* is a Banach Left A**-Bimodule

Let $\mathfrak{A}$ be a normed algebra. We will see that the second dual $\mathfrak{A}^{**}$ 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 $\mathfrak{A}^*$ can then be regarded as a Banach left $\mathfrak{A}^{**}$-module with module multiplication defined for all $\varphi \in \mathfrak{A}^{**}$ and all $f \in \mathfrak{A}^*$ via $(\varphi, f) \to \varphi f$ where $\varphi f \in \mathfrak{A}^*$ is the bounded linear functional on $\mathfrak{A}$ defined for all $a \in \mathfrak{A}$ by $(\varphi f)(a) = \varphi(fa)$ where $fa \in \mathfrak{A}^*$ is the bounded linear functional on $\mathfrak{A}$ defined for all $b \in \mathfrak{A}$ by $(fa)(b) = f(ab)$.

For each $f \in \mathfrak{A}^*$ and $a \in \mathfrak{A}$ let $fa : \mathfrak{A} \to \mathbf{F}$ be defined for all $b \in \mathfrak{A}$ by:

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

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

(5)
\begin{align} \quad (\varphi f)(a) = \varphi (fa) \end{align}
 Proposition 2: Let $\mathfrak{A}$ be a normed algebra. If $\varphi \in \mathfrak{A}^{**}$ and $f \in \mathfrak{A}^*$ then the map $\varphi f : \mathfrak{A} \to \mathbf{F}$ is a bounded linear operator
• Proof: There are two parts to this proof.
• 1. Showing that $(\varphi f)$ is linear: Let $a_1, a_2 \in \mathfrak{A}$. Then by the linearity of $\varphi$ we have that:
(6)
\begin{align} \quad (\varphi f)(a_1 + a_2) = \varphi(f[a_1 + a_2]) = \varphi (fa_1 + fa_2) = \varphi(fa_1) + \varphi(fa_2) = (\varphi f)(a_1) + (\varphi f)(a_2) \end{align}
• And for all $a \in \mathfrak{A}$ and all $\alpha \in \mathbf{F}$ we have that:
(7)
\begin{align} \quad (\varphi f)(\alpha a) = \varphi(f[\alpha a]) = \varphi(\alpha fa) = \alpha \varphi (fa) = \alpha (\varphi f)(a) \end{align}
• 2. Showing that $(\varphi f)$ is bounded. For all $a \in \mathfrak{A}$ we have that:
(8)
\begin{align} \quad \quad | (\varphi f)(a) | = | \varphi (fa) | \leq \| \varphi \| \| fa \| \overset{(\mathrm{Proposition \: 1})}\leq \| \varphi \| \| f \| \| a \| \end{align}
• So $\varphi f$ is bounded with $\| \varphi f \| \leq \| \varphi \| \| f \|$. So $\varphi f \in \mathfrak{A}^*$. $\blacksquare$

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

(9)
\begin{align} (\varphi \psi)(f) = \varphi( \psi f) \end{align}
 Proposition 3: Let $\mathfrak{A}$ be a normed algebra. If $\varphi, \psi \in \mathfrak{A}^{**}$ then the Arens product $(\varphi \psi)$ is a bounded linear operator on $\mathfrak{A}^*$.
• Proof: There are two parts to this proof.
• 1. Showing that $(\varphi \psi)$ is linear. Let $f, g \in \mathfrak{A}^*$ 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 $f \in \mathfrak{A}^*$ 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 $\varphi \psi \in \mathfrak{A}^{**}$. $\blacksquare$

And finally:

 Proposition 4: Let $\mathfrak{A}$ be a normed algebra. Then $\mathfrak{A}^{**}$ 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.