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 Nov 30 awarded Yearling Nov 30 awarded Yearling Jun 23 awarded Good Answer Feb 22 awarded Enlightened Feb 22 awarded Nice Answer Nov 30 awarded Yearling Sep 9 awarded Enlightened Sep 8 revised A property that forces the NORM to be induced by an INNER PRODUCT remove knee-jerk suggestion on second thoughts. Sep 8 revised A property that forces the NORM to be induced by an INNER PRODUCT clarify, fix typos, point out that hypothesis appears explicitly in Schoenberg. Sep 8 awarded Nice Answer Sep 7 answered A property that forces the NORM to be induced by an INNER PRODUCT Sep 7 comment A property that forces the NORM to be induced by an INNER PRODUCT See Theorem 2 of Schoenberg, A remark on M. M. Day's characterization of inner-product spaces and a conjecture of L. M. Blumenthal, Proc. Amer. Math. Soc. 3 (1952), 961-964. @Todd Trimble: I think the question is a bit more subtle than what you suggest in that the parallelogram law requires an equality rather than an inequality. Aug 28 comment Quotients of l^infty Now this is crystal clear. Thank you very much! Aug 27 comment Quotients of l^infty I'm a bit confused: doesn't Bourgain "only" construct a short exact sequence $0 \to \ell_1 \to L_1 \to X \to 0$? [This also yields the answer to the question by taking duals, using that $L_\infty$ and $\ell_\infty$ are isomorphic.] If I understand your last paragraph correctly, one can infer a short exact sequence $0 \to \ell_1 \to \ell_1 \to Y \to 0$ from this, but Bourgain doesn't seem to spell that out in his paper. Aug 26 comment A moment problem on $[0,1]$ in which infinitely many moments are equal @YemonChoi: I didn't understand that part of the answer either (I think Davide wants $+$ twice). But the following should work: Since $F(1) = 0 = m^+(1) - m^-(1)$, the measures $m^{\pm}$ have the same norm, so we can normalize to find two distinct nonzero probability measures measures whose moments-indexed-by-S agree by the choice of $F$. Aug 17 awarded Enlightened Aug 16 comment Does ZF imply a weak version of Hahn-Banach? @MohammadSafdari: You're welcome. There's surely a lot more that can be said, so feel free to wait :-) For further reading I recommend Eric Schechter's Handbook of Analysis and its Foundations for a thorough discussion of numerous weak forms of the axiom of choice and their uses in functional analysis. It is readable with only minimal background in set theory. The case of $(\ell_\infty)^\ast$ is discussed in the section on Pincus's Pathology. Aug 16 awarded Nice Answer Aug 16 answered Does ZF imply a weak version of Hahn-Banach? Jul 28 comment von neumann algebras and measurable spaces What is the precise meaning of "[the ultraweak] topology amounts to convergence in measure"? Surely you can't mean that the topologies are the same: the topology of convergence in measure is metrizable while the ultraweak topology is not metrizable except in the finite-dimensional case.