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visits | member for | 2 years |
seen | Sep 23 '13 at 3:48 | |
stats | profile views | 726 |
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. |
Jul 23 |
answered | Are dual spaces barreled? |