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S Sep 21, 2021 at 15:31 vote accept Dongyang Chen
S Sep 21, 2021 at 15:16 vote accept Dongyang Chen
S Sep 21, 2021 at 15:31
Sep 21, 2021 at 15:00 vote accept Dongyang Chen
S Sep 21, 2021 at 15:16
Sep 21, 2021 at 14:27 comment added Dongyang Chen @NarutakaOZAWA You are right. Separability of the predual is not necessary since the set of extreme points of $B_{X^{*}}$ is a James boundary of a Banach space $X$.
Sep 20, 2021 at 19:00 answer added Dirk Werner timeline score: 3
Sep 20, 2021 at 12:48 answer added Onur Oktay timeline score: 3
Sep 20, 2021 at 11:04 comment added Narutaka OZAWA @Dongyang Chen: Separability of the predual is not necessary, see Corollary 3.49. Perhaps, one can deduce it from the fact that countable subset of a compact space has isolated points??
Sep 20, 2021 at 10:03 comment added Dongyang Chen @NarutakaOZAWA Your proof uses Corollary 3.50 in mathscinet.ams.org/mathscinet-getitem?mr=1831176. But Corollary 3.50 requires that the predual of $L^{1}(\mu)$ is separable. I do not understand your proof quite well. Maybe I miss something. Thank you.
Sep 20, 2021 at 9:30 comment added Onur Oktay Building on where Professor @NarutakaOzawa has left, $X^{**}$ has Schur property. However, $X^{**}$ is a separable second dual space, so cannot have Schur property mathoverflow.net/questions/404226/…. Perhaps we may conclude that there are no biduals that are isomorphic to $L^1(\mu)$ ?
Sep 20, 2021 at 6:39 comment added Narutaka OZAWA OK, I was too haste. If $\mu$ is $\sigma$-finite, then it has countably many atoms and hence the closed unit ball of $L^1(\mu)$ has countable extreme boundary. Since $L^1(\mu)$ is assumed to be a dual Banach space, this implies that $L^1(\mu)$ is separable (Corollary 3.50 in mathscinet.ams.org/mathscinet-getitem?mr=1831176) and hence satisfies the Radon–Nikodym property. This forces $L^1(\mu)=\ell_1^n$ for $n=1,\ldots,\infty$. (I'm no Banach spacist and my proof is probably an overkill.)
Sep 20, 2021 at 3:24 history edited Dongyang Chen CC BY-SA 4.0
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Sep 20, 2021 at 3:23 comment added Dongyang Chen Thanks, Narutaka. It was known that $L^{1}[0,1]$ is not a dual space. But my question is which Banach spaces have biduals isometric to $L_{1}(\mu)$ for some $\sigma$-finite measure $\mu$. For example, the dual of every abstract M-space is abstract L-space and every abstract L-space is order isometric to $L_{1}(\mu)$ for some $\mu$.
Sep 20, 2021 at 3:02 comment added Narutaka OZAWA $L^1(\mu)$ is not even a dual Banach space, unless $\mu$ is completely atomic. This can be seen, e.g., by the fact that the closed unit ball of a dual Banach space is weak*-compact and hence has many extreme points by the Krein--Milman theorem.
Sep 20, 2021 at 2:36 history asked Dongyang Chen CC BY-SA 4.0