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 $\pi(a\otimes b)=ab$.



approximate diagonal
Unfortunately this will not work, but the counterexamples I know rely on results that are either difficult or unpublished.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 Many years ago I was told that B. E. Johnson and M. C. White had an unpublished result, observing that the corresponding algebra $B$ quotients onto the measure algebra $M({\mathbb T})$.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.


