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Let $A$ be a $C^*$-algebra with only finite dimensional irreducible representations. As in a previous question, let $\textrm{w}_0(A)$ denote the subspace of $\ell^{\infty}(A)$ consisting of all weakly null sequences.

$\textrm{w}_0(A)$ is a two-sided closed ideal of $\ell^{\infty}(A)$. Indeed, for each $f\in A^*$, $a,b\in A$, define $af(b) = f(ba) = fb(a)$. The maps $T_f,S_f:A\to A^*$ defined by $T_fa=fa$ and $S_fa=af$ are completely continuous by, e.g. Corollary 3.7 in here. Thus, if $x=(a_n)\in\textrm{w}_0(A)$ and $y=(b_n)\in\ell^{\infty}(A)$, then $$|f(a_nb_n)| \leq \|fa_n\| \|b_n\| \leq \|fa_n\| \|y\|_{\ell^{\infty}(A)}\to 0$$ for each $f\in A^*$, so $xy\in\textrm{w}_0(A)$. Similarly, $yx\in\textrm{w}_0(A)$.

Question: Is $\textrm{w}_0(A)$ an amenable Banach algebra?

note: $\textrm{w}_0(C(K))$ is amenable for $K$ a compact Hausdorff space. Indeed, $\ell^{\infty}(C(K))$ is amenable ($\ell^{\infty}(A)$ is amenable iff $A$ is subhomogeneous), and $\textrm{w}_0(C(K))$ possesses a bounded approximate identity.

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    $\begingroup$ If $A=\bigoplus_{c_0} \mathbb{M}_n$, then $\mathrm{w}_0(A)$ is not amenable (even not exact) because it contains $\prod\mathbb{M}_n$. I guess the same holds as soon as $A$ is not subhomogeneous. $\endgroup$
    – Narutaka OZAWA
    Commented Dec 19, 2023 at 6:07
  • $\begingroup$ Professor @NarutakaOZAWA, thank you for your reply. I think I get the clue. If $A$ is not subhomogeneous, then $\textrm{w}_0(A)$ contains $\Pi_{k=1}^{\infty} \mathbb{M}_{n_k}$ as subalgebra for some increasing $(n_k)$ where $n_k\to\infty$. Were $\textrm{w}_0(A)$ amenable, it would be exact, and so would its subalgebra $\Pi_{k=1}^{\infty} \mathbb{M}_{n_k}$ (which is not, I'm guessing). $\endgroup$
    – Onur Oktay
    Commented Dec 19, 2023 at 23:03
  • $\begingroup$ If $A$ is subhomogeneous, then so is $\ell^{\infty}(A)$, and so is it subalgebra $\textrm{w}_0(A)$. Subhomogeneous $C^*$-algebras are nuclear, so amenable. $\endgroup$
    – Onur Oktay
    Commented Dec 19, 2023 at 23:07

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