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Let $X$ be a Banach space. If $X^{**}$ is linearly isometric to $L_{1}(\mu)$ for some $\sigma$-finite measue $\mu$, we shall say that $X$ is an $L_{1}$-pre-bidual.

Question 1. What are the examples of $L_{1}$-pre-bidual ?

Question 2. Are there any characterizations or even references about $L_{1}$-pre-biduals ?

Thank you !

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    $\begingroup$ $L^1(\mu)$ 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. $\endgroup$
    – Narutaka OZAWA
    Commented Sep 20, 2021 at 3:02
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    $\begingroup$ 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.) $\endgroup$
    – Narutaka OZAWA
    Commented Sep 20, 2021 at 6:39
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    $\begingroup$ @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. $\endgroup$
    – Dongyang Chen
    Commented Sep 20, 2021 at 10:03
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    $\begingroup$ @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?? $\endgroup$
    – Narutaka OZAWA
    Commented Sep 20, 2021 at 11:04
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    $\begingroup$ @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 $X$. $\endgroup$
    – Dongyang Chen
    Commented Sep 21, 2021 at 14:27

2 Answers 2

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If $X$ is an infinite dimensional Banach space such that $X^{**}$ is isomorphic to $L^1(\mu)$ for some $\sigma$-finite measure, then $X^{**}$ is non-reflexive, separable and has DPP (Dunford-Pettis property) since reflexivity, separability and DPP are isomorphic properties. This is not possible (Banach spaces whose second conjugates are separable)

Edit: $L^1(\mu)$ is separable when $\mu$ is a $\sigma$-finite measure and $L^1(\mu)$ is a dual Banach space.

a. If $\mu$ is a $\sigma$-finite measure, then $L^1(\mu)$ is isometrically isomorphic to $L^1(\nu)$ for some probability measure, e.g., see Albiac & Kalton, "Topics in Banach Space Theory", Chapter 5.

b. If $\mu$ is a probability measure, then $L^1(\mu)$ is a dual space if and only if $\mu$ is purely atomic.

Proof. $(\Leftarrow)$ If $\mu$ is purely atomic, then $L^1(\mu)$ is isomorphic to $\ell^1(supp\mu)$, which is a dual space.

$(\Rightarrow)$ $L^1(\mu)$ is weakly compactly generated. Every weakly compactly generated dual Banach space has RNP, and $L^1(\mu)$ has RNP iff $\mu$ is purely atomic; see Diestel & Uhl, "Vector Measures", Section 7.7.7.

c. If $\mu$ is a finite purely atomic measure, then $supp\mu$ is countable. Thus, $\ell^1(supp\mu)$ is separable.

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    $\begingroup$ Why does it follow that $X^{**}$ is separable? (There exist $\sigma$-finite measures $\mu$ such that $L^1(\mu)$ is not separable). $\endgroup$
    – Jochen Glueck
    Commented Sep 20, 2021 at 15:33
  • $\begingroup$ @JochenGlueck This was based on the argument given above by Narutaka Ozawa. In fact, $L^1(\mu)$ (for a $\sigma$-finite $\mu$) is a dual space iff $\mu$ is purely atomic, thus $L^1(\mu) = \ell^1(supp\mu)$. Isn't $\ell^1(supp\mu)$ separable when the atomic measure $\mu$ is $\sigma$-finite? $\endgroup$
    – Onur Oktay
    Commented Sep 20, 2021 at 18:46
  • $\begingroup$ @OnurOktay The argument given by Narutaka Ozawa was based on Corollary 3.50 which requires that the predual is separable. $\endgroup$
    – Dongyang Chen
    Commented Sep 21, 2021 at 14:11
  • $\begingroup$ @OnurOktay Corollary 3.50 states that if $X$ is a separable space and the set of extreme points of the unit ball of $X^{*}$ is separable, then $X^{*}$ is separable. $\endgroup$
    – Dongyang Chen
    Commented Sep 21, 2021 at 14:16
  • $\begingroup$ @DongyangChen please see the edit. $\endgroup$
    – Onur Oktay
    Commented Sep 21, 2021 at 15:01
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If $X$ is isometric to a space $L_1(m)$, then $X^{**}$ is isometric to a (highly nonseparable) $L_1$-space over some measure space $(\Omega, \Sigma, \mu)$, by the duality of abstract $L$- and $M$-spaces. The converse is also true, as proved by Grothendieck, see Theorem II.4.9 in the Lindenstrauss-Tzafriri Springer Lecture Notes.

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  • $\begingroup$ It seems that you do not answer my question. $\endgroup$
    – Dongyang Chen
    Commented Sep 21, 2021 at 14:13
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    $\begingroup$ @DongyangChen Theorem II.4.9 gives a characterization of Banach spaces $X$ for which $X^{**}$ is an $L^1$-space. Precisely, $X^{**}$ is an $L^1$-space iff $X=L^1(\mu)$ for some measure $\mu$. Unless $X$ is finite dimensional, $X^{**}$ won't be isomorphic to $L^1(\nu)$ for a sigma-finite $\nu$. $\endgroup$
    – Onur Oktay
    Commented Sep 21, 2021 at 15:08
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    $\begingroup$ @Dongyang Chen: Apologies; I had overlooked that you are specifying $\mu$ to be $\sigma$-finite. That my answer takes care of that has been pointed out by Onur Oktay in his comment. (I have also edited my answer in that the notation for the measures was not in line with your notation.) $\endgroup$
    – Dirk Werner
    Commented Sep 22, 2021 at 9:56

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