I'm working on the amenability of some Banach algebras, and I'm wondering why $l^\infty$ is amenable ? Does any one has any idea how to start ?
Thank you in advance.
MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.
Sign up to join this communityI'm working on the amenability of some Banach algebras, and I'm wondering why $l^\infty$ is amenable ? Does any one has any idea how to start ?
Thank you in advance.
This question appears to be off-topic. The users who voted to close gave this specific reason:
In general, commutative $C^\ast$-algebras are amenable. See A new proof of the amenability of $C(X)$ by Mortaza Abtahi and Yong Zhang (Bulletin of the Australian Mathematical Society, Volume 81, Issue 3, June 2010, pages 414-417).
$l^\infty$ is a $C^*$-algebra, in fact a von Neumann algebra. For these algebras, there are many equivalent definitions of amenability. For example, for finite von Neumann algebras one such example is the existence of a Hilbert bi-module $\mathcal{H}$ and a sequence of vector $\{\xi_i\}$ such that $\|a\xi_i-\xi_ia\|_2\rightarrow 0$, for all $a$ in the algebra. You can take the usual representation of $l^\infty$ on $l^2$ by multiplication (since $l^\infty$ is a abelian this actually make it into a bimodule and the above convergence is trivial since it is abelian.