Turning Banach space into a non-trivial Banach algebra

Question

Given a Banach space, how can we determine if we can turn it into a non-trivial (i.e., not all products are zero) Banach algebra or not?

Answer 1

Well, you can turn any Banach space $A$ into a Banach algebra, by declaring all products to be $0$. Indeed, this is an algebra, since $r (x \cdot y) = (rx) \cdot y = x \cdot (ry) = 0$ and $(x+y) \cdot z = x \cdot z + y \cdot z = 0$ (here $x,y,z \in A$ and $r$ is scalar) (perhaps I omitted some condition, which but they are all trivially satisfied, with everything becoming $0$). It is associative since $(x \cdot y) \cdot z = x \cdot (y \cdot z) = 0$. This can be done in either real or complex case. The arising Banach algebra (obviously) has no unit. This is a legit Banach algebra but not a very inspiring one.

Answer 2

First of all, you should have a candidate of multiplicatiion "$\cdot$" in mind. Then, you can check the associativity of "$\cdot$", and if your Banach space is closed w.r.t. "$\cdot$". Usually, if "$\cdot$" is properly defined, associativity is not a big problem (convolution is already a rather complex case, but you can still show that.). But closeness is not that trivial. You may need to consider so-called multiplier of your Banach space w.r.t. "$\cdot$". Then the SMALLER space---all multipliers would form the pre-Banach algebra as you require (don't forget to complete it, or at least check completeness). You can refer to the paper: A note on $\tau$-convergence, $\tau$-convergent algebra and applications, Topology and its Applications 159 (2012) 1433–1438, where an example of this process is displayed. Hopefully, this might be useful to you.