Can B(l^p) ever be amenable?

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dc.creatorDaws, Matthew
dc.creatorRunde, Volker
dc.date2007-11-27
dc.date2008-06-09
dc.date.accessioned2026-07-07T09:53:52Z
dc.date.available2026-07-07T09:53:52Z
dc.descriptionIt is known that ${\cal B}(\ell^p)$ is not amenable for $p =1,2,\infty$, but whether or not ${\cal B}(\ell^p)$ is amenable for $p \in (1,\infty) \setminus \{2 \}$ is an open problem. We show that, if ${\cal B}(\ell^p)$ is amenable for $p \in (1,\infty)$, then so are $\ell^\infty({\cal B}(\ell^p))$ and $\ell^\infty({\cal K}(\ell^p))$. Moreover, if $\ell^\infty({\cal K}(\ell^p))$ is amenable so is $\ell^\infty(\mathbb{I},{\cal K}(E))$ for any index set $\mathbb I$ and for any infinite-dimensional ${\cal L}^p$-space $E$; in particular, if $\ell^\infty({\cal K}(\ell^p))$ is amenable for $p \in (1,\infty)$, then so is $\ell^\infty({\cal K}(\ell^p \oplus \ell^2))$. We show that $\ell^\infty({\cal K}(\ell^p \oplus \ell^2))$ is not amenable for $p =1,\infty$, but also that our methods fail us if $p \in (1,\infty)$. Finally, for $p \in (1,2)$ and a free ultrafilter $\cal U$ over $\posints$, we exhibit a closed left ideal of $({\cal K}(\ell^p))_{\cal U}$ lacking a right approximate identity, but enjoying a certain, very weak complementation property.
dc.description25 pages; cleaned up
dc.identifierhttps://arxiv.org/abs/0711.4311
dc.identifierhttp://arxiv.org/abs/0711.4311
dc.identifierStudia Math. 188 (2008), 151-174
dc.identifier.urihttp://salesiana.dossiersoluciones.com/handle/123456789/166105
dc.subjectFunctional Analysis
dc.subject47L10 (Primary); 46B07, 46B08, 46B45, 46E30, 46H20, 47L20 (Secondary)
dc.titleCan B(l^p) ever be amenable?
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