# 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?

• Do you mean, in which multiplication (right and left) is continuous? Jun 13, 2016 at 0:40
• @DavidHandelman : Yes sir, please forgive my carelessness Jun 13, 2016 at 1:05
• No, unit serves to ensure that the left (resp. right) regular rep. is faithful. If your Banach algebra has no zero divisor the statement works (with only one multiplication). Jun 13, 2016 at 4:09
• @DuchampGérardH.E. Can we not take the representation of $A$ on $A\oplus {\bf C}$ to get a faithful representation, and hence an equivalent norm that is submultiplicative? Jun 13, 2016 at 15:20
• @YemonChoi Of course, you're right. Jun 14, 2016 at 4:04

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.