Timeline for The Schur property and containing no isomorphic copy of $l_{1}$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 30, 2021 at 17:46 | comment | added | Tanmoy Paul | This is related to the first comment made by Narutaka; any separable Banach space can be embedded inside $\ell_\infty$. | |
| Sep 23, 2021 at 11:01 | comment | added | Dongyang Chen | Thanks, Narutaka and Onur. You are right. | |
| Sep 23, 2021 at 10:50 | comment | added | Onur Oktay | A particular case of Pełczyński's result can also be found in Corollary A in arxiv.org/abs/2108.03057v1. | |
| Sep 23, 2021 at 10:39 | history | edited | LSpice | CC BY-SA 4.0 |
Deleting spaces before punctuation
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| Sep 23, 2021 at 10:18 | comment | added | Onur Oktay | @DongyangChen If a Banach space $Y$ contains a copy of $\ell^1$, then $Y^{*}$ contains a copy of the space of measures $M([0,1]) = C([0,1]) ^{*}$, which contains a copy of $\ell^1([0,1])$. This was proven by Pełczyński, see Proposition 3.3 in https:/doi.org/10.4064/sm-30-2-231-246 | |
| Sep 23, 2021 at 9:17 | comment | added | Dongyang Chen | I do not know $X^{**}$ would contain $l_{1}$ whenever $X^{*}$ contains $l_{1}$. Could you give a detailed proof ? | |
| Sep 23, 2021 at 9:12 | comment | added | Narutaka OZAWA | According to what you said, $X^*$ contains $\ell_1$, which obviously implies $X^{**}$ contains $\ell_1$ as well. | |
| Sep 23, 2021 at 8:03 | history | asked | Dongyang Chen | CC BY-SA 4.0 |