977 reputation
616
bio website
location
age
visits member for 1 year, 8 months
seen Sep 23 '13 at 3:48

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?
Jul
20
comment Origin and first uses of $\ell_p$ norms?
I believe Riesz's reference to page 95 of Minkowski is a misprint for page 85 where Minkowski shows that convexity of a solid containing the origin is equivalent to homogeneity and the triangle inequality of its associated gauge functional (equations (37), (39)). The convexity of the two-dimensional $\ell_p$-balls is proved on page 48, where there is the familiar picture of the nested $\ell_p$ unit balls in the plane.