A unital algebra with norm and continuous multiplication is a Banach algebra In my research in functional analysis, I came across this rather simple result:

For a normed algebra A over $ \mathbb{C} $ with unit, in which multiplication , right and left are both continuous w.r.t the norm defined, such that A with this norm is a Banach space, there exists an equivalent norm with respect to which A is a Banach Algebra.

I was wondering, is the existence of a unit in A truly necessary, or can we guarantee the existence of an equivalent norm such that A is a Banach algebra? 
 A: The answer is no (a unit is not necessary within the algebra). 
In the case the product is continuous (i.e. there exists $M>0$ such that, identically, $||xy||\leq M||x||.||y||$ see this question) and of a unital algebra, the common way to construct a multiplicative norm equivalent to the given one is to consider the left-regular representation i.e. $s\rightarrow \gamma(s)$ where $\gamma(s)\in \mathcal{L}_A$ is defined by $\gamma(s)[x]=sx$ and setting the new norm $||s||':=|||\gamma(s)|||$ (the last norm being that of the bounded convergence within $\mathcal{L}_A$). In this respect, the representation $s\rightarrow \gamma(s)$ must be faithful (it is the case, in particular, when $A$ is unital but not only). If it is not, then $||\ ||_1$ is only a seminorm. Following Yemon's comment, in the case when the given algebra is not unital, a way to circumvent this is to consider the left-regular $s\rightarrow \gamma_1(s)$ representation of $A$ on the Banach space $A\oplus\mathbb{C}$ (endowed, for example, with the norm $||x+\lambda||_1=||x||+|\lambda|$). This representation is given by $\gamma_1(s)[x+\lambda]=sx+\lambda.s$. It is an easy exercise to prove that the new seminorm $||s||'':=|||\gamma_1(s)|||$ is, in any case, a norm equivalent to the given one. Hope that it helps.   
