# 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)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. Showing that $(fa)$ is bounded.**For all $b \in \mathfrak{A}$ we have by the definition of $fa$ and the boundedness of $f$ that:

- 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)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:

- And for all $a \in \mathfrak{A}$ and all $\alpha \in \mathbf{F}$ we have that:

**2. Showing that $(\varphi f)$ is bounded.**For all $a \in \mathfrak{A}$ we have that:

- 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:

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:

**2: Showing that $\varphi \psi$ is bounded.**For all $f \in \mathfrak{A}^*$ we have that:

- 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.*