Timeline for What is the Banach-Mazur distance between $\ell_\infty$ and $L_\infty$?
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13 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jun 9, 2016 at 15:34 | comment | added | Bunyamin Sari | @YemonChoi Arbitrarily big, e.g., the distance between $C(\omega)$ and $C(\omega^n)$ tends to infinity as $n$ does. | |
| Jun 9, 2016 at 15:26 | comment | added | Yemon Choi | @BillJohnson If $C(K_1)\cong C(K_2)$, how big can the BM-distance between them be? That is, do we have examples to show that the BM-distance can be arbitrarily big? | |
| Jun 9, 2016 at 6:41 | comment | added | Bill Johnson | Another way to express Robert Israel's comment is to say that the Stone space of $\ell_\infty$ has isolated points while the Stone space of $L_\infty$ does not. | |
| Jun 9, 2016 at 6:37 | comment | added | Bill Johnson | It is not true, Fedor. Take a sequence $p_n$ that strictly decreases to $1$ and consider $X= (\sum_{n=1}^\infty \ell_{p_n}^2)_2$. Then the BM distance of $X$ to $X\oplus_2 \ell_1^2$ is one but the spaces are not isometrically isomorphic because the latter one is not strictly convex. | |
| Jun 9, 2016 at 6:18 | comment | added | Fedor Petrov | By the way, how to prove (if it is true in general, not only for $C(K)$ spaces) that whenever two spaces are not isometric, BM distance is strictly greater than 1? | |
| Jun 9, 2016 at 6:12 | comment | added | Fedor Petrov | @AnthonyQuas this lattice is defined using norm and linear structures only, thus Robert's argument looks fine. | |
| Jun 9, 2016 at 5:56 | comment | added | Anthony Quas | Doesn't this prove just that there's no lattice-preserving isometry from $\ell_\infty$ to $L_\infty$? | |
| Jun 9, 2016 at 0:40 | comment | added | Robert Israel | Choose an extreme point of the unit ball, and call it $e$. We can define a partial ordering on our space by $x \ge 0$ if $\|\|x\| e - x \| \le \|x\|$. This makes $L_\infty$ or $\ell_\infty$ into a Banach lattice (equivalent by an isometry to the usual lattice structure). Now $\ell^\infty$ has the property that it has minimal nonnegative elements of norm $1$, i.e. $x$ such that the only element $0 \le y \le x$ with $\|y\| = 1$ is $x$. But $L_\infty$ does not have such elements. So $\ell_\infty$ and $L_\infty$ are not isometric. | |
| Jun 9, 2016 at 0:31 | history | edited | Bunyamin Sari | CC BY-SA 3.0 |
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| Jun 9, 2016 at 0:14 | comment | added | Bunyamin Sari | Thanks for this. I will edit it once someone points out a correct argument for the non-isometry of these spaces. | |
| Jun 9, 2016 at 0:05 | comment | added | Anthony Quas | I don't think it is right that the unit ball of $L_\infty$ has more extreme points than the unit ball of $\ell_\infty$. Here's the argument in the real case. A point in the unit ball of $L_\infty$ is $2\mathbf 1_B-1$, where $B$ is a measurable set. Two such functions agree if the $B$'s differ by a set of measure 0. Any Lebesgue measurable $B$ agrees with a Borel $B$ up to a set of measure 0. The cardinality of the Borel $\sigma$-algebra on $[0,1]$ is $\aleph_1$, the same as the cardinality of the collection of $\pm 1$ sequences. | |
| Jun 8, 2016 at 23:52 | history | answered | Bunyamin Sari | CC BY-SA 3.0 |