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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?

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    $\begingroup$ Do you mean, in which multiplication (right and left) is continuous? $\endgroup$
    – David Handelman
    Commented Jun 13, 2016 at 0:40
  • $\begingroup$ @DavidHandelman : Yes sir, please forgive my carelessness $\endgroup$
    – Don John Prep
    Commented Jun 13, 2016 at 1:05
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    $\begingroup$ 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). $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 13, 2016 at 4:09
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    $\begingroup$ @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? $\endgroup$
    – Yemon Choi
    Commented Jun 13, 2016 at 15:20
  • $\begingroup$ @YemonChoi Of course, you're right. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 14, 2016 at 4:04

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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.

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  • $\begingroup$ I don't understand: I thought the answer to the given question is yes. Certainly we in BA world believe that if one has a Banach space, equipped with an algebra product that is continous, then this space can be renormed with an equivalent norm such that one gets a Banach algebra in the usual sense. $\endgroup$
    – Yemon Choi
    Commented Jun 17, 2016 at 21:17
  • $\begingroup$ @Yemon Choi: I read it this way: The OP asks if a unit is necessary for the existence of an equivalent multiplicative norm and Duchamp Gérard H. E.s answer says "No, a unit is not necessary - such norm always exists". $\endgroup$
    – Todd Leason
    Commented Jun 18, 2016 at 8:57
  • $\begingroup$ @ToddLeason I see what you mean. Thanks $\endgroup$
    – Yemon Choi
    Commented Jun 18, 2016 at 14:26
  • $\begingroup$ @YemonChoi Yes, Todd is right in his interpretation of my ambiguous "no". I made it precise. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 18, 2016 at 16:19
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    $\begingroup$ In this and in many other cases on OM, a simple answer y/n is not easy to interpret. There are still many other answers that no answer in the "Answer" is formulated. I feel that an answer should alway include the full statement of its result explicitly. Preferably, most of the time, an answer should start with the explicit full statement of its result. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Jul 10, 2016 at 18:16

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