# Amenability of $l^\infty$ [closed]

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 ?

## closed as off-topic by Nik Weaver, Chris Godsil, Yemon Choi, Andrey Rekalo, Ricardo AndradeApr 3 '14 at 1:37

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• As indicated in the answer below: see Johnson's 1972 monograph, or the proof in Bonsall and Duncan's book – Yemon Choi Mar 30 '14 at 4:46
• This question admits an easy answer if you know the right references, so I am voting to close – Yemon Choi Mar 30 '14 at 17:15

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).
• Can we prove it using the definition only ? Can we prove the existence of a bouded approximate diagonal for $l^\infty$ – user128591 Mar 30 '14 at 4:48
$l^\infty$ 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.