# example of the arens multiplication; I want to understand the construction

I want to understand the double dual as a $C^*$-algebra of a given $C^*-$algebra $A$ but my first problem is to understand the Arens multiplication on the double dual $A^{**}$ (considered as Banach space) of $A$, it is defined here https://www.encyclopediaofmath.org/index.php/Arens_multiplication for example (if you want that I repeat the definition here, I will do it). There are 2 Arens multiplications but they coincide if you consider $C^*$-algebras.

Maybe I will understand the definition if I see the construction/the definition of the product demonstrated on an example. For example I could take the $C^*$-algebra $c_0=\{ (a_n)_{n\in\mathbb{N}}\subseteq \mathbb{C}; \lim_{n\to\infty}a_n=0 \}$ endowed with the pointwise multiplication, the maximum norm, and $*:c_0\to c_0,\; (a_n)\mapsto (a_n)^*=(\overline{a_n})$ as involution. It's dual space can be identified with $l^1$ and the dual space of $l^1$ can be identified with $l^{\infty}$, but $c_0$ isn't reflexive. But via the canonical embedding $i: c_0\to l^{\infty},\; (a_n)\mapsto i(a_n)$ cou can identify $c_0$ as a subalgebra of $l^{\infty}$ and the Arens multiplication should correspond to the usual pointwise multiplication in $l^{\infty}$. But if I see the construction of the Arens product I don't see it. I want to understand the construction of the Arens product, could you explain me this? (If you want we could take an other example). Regards

All you have to keep in mind are the natural identification of $\ell^1$ and $c_0^\ast$ resp. $\ell^\infty$ and $(\ell^1)^\ast$. I will write $\ast$ for the product in the three steps of the construction of the Arens product so that there is no confusion with pointwise multiplication.
For $a,b\in c_0, \omega\in\ell^1$ we have $$\langle a\ast\omega,b\rangle=\langle\omega,ab\rangle=\sum_{n=1}^\infty \omega(n)a(n)b(n)=\sum_{n=1}^\infty (a\omega)(n)b(n)=\langle a\omega,b\rangle.$$
Essentially the same computation yields $f\ast\omega=f\omega$ and $f\ast g=fg$ for all $\omega\in\ell^1,f,g\in\ell^\infty$ (just carefully write down the definitions).
By the way, the universal (GNS-) representation $\pi$ of a $C^\ast$-algebra $A$ extends to an isomorphism $A^{\ast\ast}\longrightarrow \pi(A)''$. In the case of $A=c_0$, the GNS Hilbert space is just $\ell^2$ and $c_0$ is represented as multiplication operators. It is easy to see that $\pi(A)''\cong \ell^\infty$ (also as multiplication operators) and the composition of multiplication operators is just given as pointwise multiplication of the associated functions.