Timeline for Is there a criterion for compactness in $L^\infty(\Omega)$ with strong topology?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2017 at 8:25 | answer | added | don | timeline score: 5 | |
| Sep 13, 2017 at 14:28 | comment | added | Nate Eldredge | Well, a simpler example is $[0,2]$ versus $[0,1] \cup (2,3]$; just split $[0,2]$ in half and move one half over. To get to $[2,3]$ instead of $(2,3]$, take some countable set $C = \{x_1, x_2, x_3, \dots\}$ in $(2,3]$, let $x_0 = 2$, and consider the map that fixes $C^c$ and maps $x_n$ to $x_{n-1}$. | |
| Sep 13, 2017 at 14:04 | comment | added | Ye Changqing | Thanks again, but how to construct a bi-measurable function which takes null sets to null sets? | |
| Sep 13, 2017 at 13:36 | comment | added | Nate Eldredge | Topological isomorphism is neither necessary or sufficient. What you need is a bimeasurable map that takes null sets to null sets. If $\Omega$ is the standard measure-zero Cantor set and $\Omega'$ is a positive-measure "fat" Cantor set, then they are homeomorphic but $L^\infty(\Omega) = 0$ while $L^\infty(\Omega')$ is infinite dimensional. Conversely, $\Omega = [0,1]$ and $\Omega' = [0,1] \cup [2,3]$ are not homeomorphic but they have the same $L^\infty$. | |
| Sep 13, 2017 at 13:26 | comment | added | Ye Changqing | Thanks, but why all $L^\infty(\Omega)$ are isomorphic? if $\Omega_1$ and $\Omega_2$ are topologically isomorphic, the isomorphism between $L^\infty(\Omega_1)$ and $L^\infty(\Omega_2)$ is trivial. P.s. the criterion in Dunford and Schwartz's book seems very hard to apply, sad... | |
| Sep 13, 2017 at 4:52 | comment | added | Nate Eldredge | See math.stackexchange.com/questions/955394/…. Surprise surprise: the place to look is Dunford and Schwartz. | |
| Sep 13, 2017 at 4:49 | comment | added | Nate Eldredge | Compactness of $\Omega$ can't be relevant here - $L^\infty(\Omega)$ only depends on the measure algebra of $\Omega$, and that's the same for every positive-measure set $\Omega$, so all $L^\infty(\Omega)$ are isomorphic (except the trivial ones where it is zero-dimensional). | |
| Sep 13, 2017 at 3:59 | review | First posts | |||
| Sep 13, 2017 at 4:06 | |||||
| Sep 13, 2017 at 3:58 | history | asked | Ye Changqing | CC BY-SA 3.0 |