Let $A$ be a $C^\ast$-algebra. Consider the canonical embedding $A\to A^{\*\*},\; a\mapsto i(a)$, such that $i(a)(a^\*)=a^\*(a)$ for all $a\in A$. Here is $A^{\*\*}$ considered as a Banach space. It's well known that $i(A)$ is weak$\*$-dense in $A^{\*\*}$. I want to endow $A^{\*\*}$ with an inner multiplication, a norm and an involution map, such that $A^{\*\*}$ is a $C^\ast$-algebra. The canonical embedding $i$ is not multiplicative in general. Therefore I want to know: Does it make sense, if one defines the multiplication and an involution on $A$ as follows: $i(a)i(b):=i(ab)$ for all $a,b\in A$ and $i(a)^\*:=i(a^\*)$ for all $a\in A$ and then you extend continuously on $A^{\*\*}$? 


Regards



short question about biduals of $C^\ast$-algebras
I'm not sure what you mean by "the canonical embedding is not multiplicative" if you're only taking $A^{\*\*}$ as a Banach space. A natural way to give $A^{\*\*}$ multiplicative structure is via the [Arens product](http://www.encyclopediaofmath.org/index.php/Arens_multiplication), which does make the canonical embedding multiplicative. So this shows that your prescription of importing a product into $A^{\*\*}$ using $i$ and extending by continuity does work. The analogous statements for the adjoint operation are also easy.


Since you're asking about the C\*-algebra structure of $A^{\*\*}$, are you aware that it is naturally $\*$-isomorphic to the weak\* completion of $A$ in its universal representation? This construction is called the [universal enveloping von Neumann algebra](https://en.wikipedia.org/wiki/Enveloping_von_Neumann_algebra) and the statement of isometry is the [Sherman-Takeda theorem](https://en.wikipedia.org/wiki/Sherman%E2%80%93Takeda_theorem).


