Let $v=(v_n)_{n\in\mathbb{N}}$ be a positive weight with $\inf_nv_n>0$ (for convenience we may take $v_n\geqslant1$). Then $\ell_{\infty}(v)$ is a Banach algebra with coordinate-wise multiplication. Is it weakly amenable?
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$\begingroup$ Isn't $\ell_\infty(v)$ a von Numann algebra? All von Numann algebras are weakly amenable. $\endgroup$– MSMalekanCommented May 4, 2018 at 18:32
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$\begingroup$ @MeisamSoleimaniMalekan It isn't a von Neumann algebra because it has a different norm. $\endgroup$– Yemon ChoiCommented May 6, 2018 at 18:54
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1$\begingroup$ To the OP: can you clarify if your norm is defined by $\sup_n |x_n| \nu_n$ or $\sup_n |x_n| \nu_n^{-1}$? $\endgroup$– Yemon ChoiCommented May 6, 2018 at 18:54
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1$\begingroup$ @Norbert, if $(v_n)_n$ is bounded then $\ell_{\infty}(v)$ is isomorphic (as a Banach algebra) to $\ell_{\infty}$ (recall $\inf_nv_n>0$) and so even amenable. Assume now $v$ is an unbounded weight. Wlog let $v$ diverge to infinity. Consider (this is a result of my discussion with Mateusz Wasilewski and Mikolaj Fraczyk) a functional $\phi$ on $\ell_{\infty}(v)$ defined as $\phi(x):=\lim_{\mathcal{U}}x_nv_n$ over a free ultrafiler $\mathcal{U}$ on $\mathbb{N}$. It is continuous, it vanishes on $\overline{\ell_{\infty}(v)^2}$ and $\phi(v^{-1})=1$. $\endgroup$– KrzysztofCommented Jul 2, 2018 at 21:49
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1$\begingroup$ Thus $v^{-1}\in\ell_{\infty}(v)\setminus\overline{\ell_{\infty}(v)^2}$. This means that the algebra $\ell_{\infty}(v)$ is not essential and this is a weaker condition than weak amenability -- see Theorem 2.8.63 in Dales' monograph "Banach Algebras and Automatic Continuity". Consequently, if $\ell_{\infty}(v)$ is an algebra then it is either amenable or not weakly amenable. $\endgroup$– KrzysztofCommented Jul 2, 2018 at 21:54
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