# 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)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. Showing that $(fx)$ is bounded.**For all $y \in X$ we have by the definition of $fx$ and the boundedness of $f$ that:

- 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)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:

- (The equality at $(*)$ comes from the fact that $X$ is an ideal of $X$ and so by the Examples of Modules - Left Ideals J are Left X-Modules page we have that $X$ in particular is a normed left $X$-module. So by the Examples of Modules - For Normed Algebras X and Left X-Modules M, M* is a Banach Right X-Module we have that $X^*$ is a Banach right $X$-module. The equality at $(**)$ comes from the fact that $\varphi$ is linear functional on $X^*$.)

- Now for all $x \in X$ and all $\alpha \in \mathbf{F}$ we have that:

**2. Showing that $(\varphi f)$ is bounded.**For all $x \in X$ we have that:

- 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:

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:

**2: Showing that $\varphi \psi$ is bounded.**For all $f \in X^*$ we have that:

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

(14)Using the above we see that:

(15)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.