approximate diagonal

Question

Let $I$ be an arbitary index set, $((A_i)_i,\|.\|_i)_{i\in I}$ be a family of Banach algebras, with approximate diagonal $(m^{i}_α)_α\subseteq A_i\hat\otimes A_i$, and $B=\{(x_i)_i\in {\displaystyle \prod _{i\in I}}A_i|\, \sum_i\|x_i\|_i<\infty\}$. Is there any approximate diagonal for $B$?

My idea: For finite subset ‎‎$‎‎‎‎F‎‎\subseteq I‎$‎ we define ‎$‎E_F‎$‎‎‎ as below ‎‎‎‎‎ ‎
‎$$ (‎E_F)_‎{i}=\Big\{^{m_\alpha^{i}~~{~i\in F}}_{0~~\mbox{‎ else. ‎}} $$‎‎ Then $(‎E_F)_{F}$ is an approximate diagonal for $B$, ‏where partial order is defined as ‎‎‎‎‎$F_1‎\preceq F_2$ ‏if and only if $ ‎F_1\subseteq F_2‎$‎‎‎‎. Am I right? If yes, how can I prove that $\pi((‎E_F)_‎{i})a-a\to 0$ in norm?

Here, $\hat{\otimes}$ denotes the projective tensor product of Banach spaces and $\pi\colon B\hat\otimes B\to B$ is defined by $\pi(a\otimes b)=ab$.

Answer 1

Unfortunately this will not work, but the counterexamples I know rely on results that are either difficult or unpublished.

1) Take $A_n = M_n{\mathbb C})$ with usual multiplication and $C^*$-norm. Each $A_n$ is amenable with constant $1$, so there exists an actual diagonal element in $A_n\hat\otimes A_n$ which has norm $1$. On the other hand, your algebra $B$ is an example of a non-nuclear $C^*$-algebra and hence it is non-amenable.

2) Take $A_n = \ell^1({\mathbb T}_d)$ where ${\mathbb T}_d$ denotes the circle group equipped with the discrete topology. Once again each $A_n$ is amenable with constant $1$. Many years ago I was told that B. E. Johnson and M. C. White had an unpublished result, observing that the corresponding algebra $B$ quotients onto the measure algebra $M({\mathbb T})$. Now it is is known by work of Brown and Moran that $M({\mathbb T})$ has a non-zero point derivation, and pulling this back to $B$ we see that $B$ is not weakly amenable, hence it is not approximately amenable.

(I think example 2 is actually mentioned briefly near the end of Ghahramani and Loy's first paper on "generalized notions of amenability", but I haven't had time to check this.)

Relatedly: knowing that a Banach algebra $A$ is amenable does not guarantee that $\ell^\infty({\mathbb N}; A)$ is amenable. For instance one can take $A={\mathcal K}(\ell_p)$, known to be amenable by work in Johnson's 1972 memoir; and then Runde has shown that $\ell^\infty({\mathbb N}; {\mathcal K}(\ell_p) )$ is non-amenable.