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It was known that the James space $J$ has separable second conjugate, is non-reflexive and isometric to its second conjugate. I want to know whether there are Banach spaces $X$ with separable second conjugates $X^{**}$, but $X$ is not a dual space (the James space $J$ is a dual space). Furthermore, are there any references about Banach spaces with separable second conjuates ?

Thank you !

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    $\begingroup$ Is "second conjugate" the same as "bidual"? $\endgroup$
    – YCor
    Commented Sep 18, 2021 at 7:54
  • $\begingroup$ @YCor Yes, it is. $\endgroup$ Commented Sep 18, 2021 at 9:35

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Yes, there are such spaces. To see this, first note that Joram Lindenstrauss showed that for every separable Banach space $Y$ there exists a Banach space $X$ such that $X^{\ast\ast}$ is separable and $X^{\ast\ast}/X$ is isomorphic to $Y$. Apply this result with $Y$ a separable Banach space that is not isomorphic to a subspace of a separable dual space (e.g., $c_0$ or $L_1$; see for example Theorem 6.3.7 from Albiac and Kalton's book Topics in Banach Space Theory for details). Then, for such $Y$, the space $X$ prescribed by Lindenstrauss' construction cannot be isomorphic to a dual space. Indeed, if it were, then $X^{\ast\ast}/X$ would be isomorphic to a subspace of $X^{\ast\ast}$ by virtue of the fact that that every dual space is complemented in its bidual (in particular, $X^{\ast\ast}$ would be isomorphic to $(X^{\ast\ast}/X)\oplus X$), and thus $Y$ would also be isomorphic to a (complemented) subspace of the separable dual space $X^{\ast\ast}$ - a contradiction.

(As it happens, the aforementioned book of Albiac and Kalton includes Lindenstrauss' result in the chapter with the title 'Important Examples of Banach Spaces'.)

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This post is intended not as an answer, but rather to list several Banach space properties of $X$ given above (that is $X^{**}$ is separable, $X$ is not a dual space).

  1. Clearly $X$ is not reflexive.
  2. Clearly $X$ and $X^{*}$ are also separable.
  3. Being separable dual spaces, $X^{*}$ and $X^{**}$ have RNP (Radon-Nikodym property). $X$ also has RNP, being a closed subspace of a space with RNP.
  4. $X$, $X^{*}$ do not contain copies of $\ell^1$.
  5. $X$, $X^{*}$, $X^{**}$ do not contain copies of $c_0$.
  6. $X$, $X^{*}$ do not possess an unconditional basis.
  7. $X$, $X^{*}$ are not w.s.c. (weakly sequentially complete) since any Banach space which is w.s.c. and contains no copy of $\ell^1$ is reflexive.
  8. $X$, $X^{*}$ do not have Schur property, for any Banach space that has Schur property and contains no copy of $\ell^1$ is finite dimensional.
  9. $X$ does not have DPP (Dunford-Pettis property), since otherwise $X^{*}$ would have Schur property. Consequently, $X^{*}$ and $X^{**}$ do not have DPP (and thus not have Schur property)
  10. $X$, $X^{*}$, $X^{**}$ do not have Pełczyński property (V), for any Banach space that has property (V) and contains no copy of $c_0$ is reflexive.
  11. $X$, $X^{*}$ have Dieudonne property, for any Banach space that contains no copy of $\ell^1$ has Dieudonne property.
  12. $X$, $X^{*}$, $X^{**}$ are not Grothendieck spaces, since every separable Grothendieck space is reflexive.

The properties above are also shared by the James space, since we merely used that "$X$ is not a dual space" to deduce that $X$ is not reflexive.

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    $\begingroup$ Your listed properties of the Banach space $X$ are extensive. $\endgroup$ Commented Sep 20, 2021 at 1:51
  • $\begingroup$ Could you give proofs of 4 and 5 ? $\endgroup$ Commented Sep 20, 2021 at 9:54
  • $\begingroup$ @DongyangChen (4) is due to separability. If a Banach space $Y$ contains a copy of $\ell^1$, then $Y^{*}$ contains a copy of the measure space $M([0,1])$, which is not separable. Thus, $X$, $X^{*}$ do not contain copies of $\ell^1$. (5) If $Y^{*}$ contains a copy of $c_0$, then $Y$ contains a complemented copy of $\ell^1$. Thus, $X^{*}$, $X^{**}$ do not contain copies of $c_0$. And thus, $X\subset X^{**}$ does not contain a copy of $c_0$. $\endgroup$
    – Onur Oktay
    Commented Sep 20, 2021 at 10:29
  • $\begingroup$ (4) also follows from (3). $\endgroup$
    – Onur Oktay
    Commented Sep 20, 2021 at 10:31

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