# The First and Second Arens Products on A**

## The First Arens Product

Let $\mathfrak{A}$ be a Banach algebra. Consider the second dual, $\mathfrak{A}^{**}$, which is clearly a Banach space. We would like to make $\mathfrak{A}^{**}$ a Banach algebra too, but it is not entirely obvious what the multiplication on $\mathfrak{A}^{**}$ should be. One type of multiplication we can define on $\mathfrak{A}^{**}$ is the **First Arens Product on $\mathfrak{A}^{**}$**. It is defined in steps as follows:

Definition: Let $\mathfrak{A}$ be a Banach algebra.1. For each $a \in \mathfrak{A}$ and for each $f \in \mathfrak{A}^*$ we define $f \cdot a \in \mathfrak{A}^*$ by $(fa)(b) := f(ab) \quad (\forall b \in \mathfrak{A})$. 2. For each $F \in \mathfrak{A}^{**}$ and for each $f \in \mathfrak{A}^*$ we define $F \cdot f \in \mathfrak{A}^*$ by $(F \cdot f)(a) := F(fa) \quad (\forall a \in \mathfrak{A})$ 3. Lastly, the First Arens Product on $\mathfrak{A}^{**}$ is the multiplication on $\mathfrak{A}^{**}$ defined for all $F, G \in \mathfrak{A}^{**}$ by $(FG)(f) := F(Gf) \quad (\forall f \in \mathfrak{A}^*)$. |

It is clear that the First Arens Product defined above satisfies the 3 axioms on the Algebras over F page, making $\mathfrak{A}^{**}$ with the First Arens Product a Banach algebra. Indeed, let's verify these three axioms

**1)**Let $F, G, H \in \mathfrak{A}^{**}$. We will show that $[FG]H = F[GH]$. This is done by showing that these functionals are equal for all $f \in \mathfrak{A}^*$. Indeed:

**2)**Let $F, G, H \in \mathfrak{A}^{**}$. We now show that $F[G + H] = FG + FH$. This is again done by showing that these functionals are equal for all $f \in \mathfrak{A}^*$. Indeed by the linearity of $F$ we have that:

**3)**Let $F, G \in \mathfrak{A}^{**}$ and let $\alpha \in \mathbb{C}$. We lastly show that $[\alpha F]G = \alpha [FG] = F[\alpha G]$. Yet again, this is done by showing that these functionals are equal for all $f \in \mathfrak{A}^*$:

- So that $[\alpha F]G = \alpha [FG]$. Also:

- So that $\alpha [FG] = F[\alpha G]$ too.

## The Second Arens Product

We can quite naturally define another type of multiplication on $\mathfrak{A}^{**}$ call the **Second Arens Product** as follows:

Definition: Let $\mathfrak{A}$ be a Banach algebra.1) For each $a \in \mathfrak{A}$ and for each $f \in \mathfrak{A}^*$ we define $af \in \mathfrak{A}^*$ by $(af)(b) := f(ba) \quad (\forall b \in \mathfrak{A})$.2) For each $F \in \mathfrak{A}^{**}$ and for each $f \in \mathfrak{A}^*$ we define $fF \in \mathfrak{A}^*$ by $(fF)(a) := F(af) \quad (\forall a \in \mathfrak{A})$.3) Lastly, the Second Arens Product on $\mathfrak{A}^{**}$ is the multiplication on $\mathfrak{A}^{**}$ defined for all $F, G \in \mathfrak{A}^{**}$ by $(F * G)(f) :=F(fG) \quad (\forall f \in \mathfrak{A}^*)$. |

*Here we will use $*$ to explicitly denote the second Arens product.*

## Equality of the First and Second Arens Products

In general, given $F, G \in \mathfrak{A}^{**}$ it may be that $FG$ (first Arens product) is not equal to $F * G$ (second Arens product). When these two product are equal for all $F, G \in \mathfrak{A}^{**}$, the Banach algebra $\mathfrak{A}$ is given a special name.

Definition: Let $\mathfrak{A}$ be a Banach algebra. Then $\mathfrak{A}$ is said to be Arens Regular if $FG = F*G$ for all $F, G \in \mathfrak{A}^{**}$, where the lefthand side of the equality is multiplication with respect to the first Arens product, and the righthand side of the equality is multiplication with respect to the second Arens product. |