# Banach spaces whose second conjugates are separable

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 !

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

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'.)

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.

• Your listed properties of the Banach space $X$ are extensive. Sep 20, 2021 at 1:51
• Could you give proofs of 4 and 5 ? Sep 20, 2021 at 9:54
• @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 $c_0$. Sep 20, 2021 at 10:29
• (4) also follows from (3). Sep 20, 2021 at 10:31