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
 A: A late answer, but maybe still helpful.
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.
