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Let us have a look on the proof of Theorem 2 in [P. G. Spain, Boolean algebras of projections, Proceedings of the Edinburgh Mathematical Society (Series 2) 19, 03, March 1975, 287-289]

The author claims in the proof of Theorem 2 that if $B$ is a (Bade) complete Boolean algebra of projections on a Banach space which is relatively weakly compact subset of the algebra of operators, then the weak (WOT?) closure of the algebra generated by $B$ is a W*-algebra.

I don't think the proof is OK. Take $p\neq 2$, $p\in (1,\infty)$ and consider $X=\ell_p$ with its canonical basis, which is 1-unconditional. The family $B$ of basis projections forms a complete Boolean algebra (actually isomorphic to $\wp(\omega)$) and is relatively weakly compact because all the projections have norm one and the unit ball of the space of operators on a reflexive space is WOT-compact. Though, $\overline{\mbox{alg }B}^{WOT}=B(\ell_p)$ which is not a von Neumann algebra (it is not even Banach-space isomorphic to a von Neumann algebra).

My question is (before it's closed by the math police). Is Theorem 2 correct?

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  • $\begingroup$ Not sure this is a question- voting to close. $\endgroup$
    – Daniel Moskovich
    Commented Feb 18, 2013 at 13:49
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    $\begingroup$ If this turns into a protracted tug-of-war then of course the discussion should be moved to meta. But, for the record, it seems to me that this is a real question. If anything, the OP suffers from having thought too productively about it, by producing a (putative) counterexample. Without that, Spain's claim could just be rephrased as a question. $\endgroup$
    – HJRW
    Commented Feb 18, 2013 at 15:19
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    $\begingroup$ Why not email Philip Spain first? His email is available on his homepage: maths.gla.ac.uk/~pgs $\endgroup$
    – Dmitri Pavlov
    Commented Feb 18, 2013 at 15:42
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    $\begingroup$ I find mention of the "math police" tiresome and unhelpful. $\endgroup$
    – Yemon Choi
    Commented Feb 18, 2013 at 19:45
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    $\begingroup$ Also, having had a quick look at Theorem 2, I can't see where there might be a mistake. So I would respectfully venture that the answer to your question is YES. $\endgroup$
    – Yemon Choi
    Commented Feb 18, 2013 at 19:51

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I haven't looked at the paper but your counterexample is mistaken. The basis projections generate not $B(l^p)$ but the algebra of multiplication operators, which is isometrically isomorphic to $l^\infty$ and hence is a von Neumann algebra.

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