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Let $Y$ be a complemented, but not $w^{*}$-closed, subspace of a Banach space $X$. It is known that certain such $Y$ are not dual spaces.

Question: What are interesting examples of subspaces of the above type that are dual spaces?

The example linked to above tells us that a subspace being complemented in a dual space gives no evidence that the subspace is a dual space. The intent of this question is really to sharpen this intuition by looking at examples. For instance, these examples may reveal a special way to tell whether such a subspace is a dual space refining general facts found here. Taking this reference together with the negative example above, one might conclude that one cannot readily refine the predual of the big space to obtain a predual for the subspace, but this doesn't block the subspace having some completely unrelated predual. The general intent of this question is to find examples that show if, in certain situations, the predual is ``somewhat related'' to the predual of the larger space.

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    $\begingroup$ I'm writing down the first example that comes to my mind, so apologies in advance for probable mistakes: Let $D\subset [0,1]$ be the set of dyadic numbers, $\delta_D = \{\delta_r: r\in D\}\subset M([0,1])$, $Y$ be the closed linear span of $\delta_D$. It's not difficult to verify that $Y$ is isomorphic to $\ell^1$, $Y$ is complemented in $M([0,1])$, and $Y$ is not weak$^*$ closed. $\endgroup$
    – Onur Oktay
    Commented Jun 11, 2023 at 13:58
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    $\begingroup$ I'm not an MO editor but I believe editing/closing the question is your call. I'm glad if you found this example useful. It isn't exactly a home run - perhaps as useful as pepper. As an attempt for a pitch, let $A$ be a $C^*$-algebra that contains a complemented $\ell^2$ subspace and $M=A^{**}$. This is the case whenever $A^{**}$ is not type I finite. Then $M$ also contains a complemented $\ell^2$ subspace. I'm not sure at the moment if this subspace is always weak$^*$ closed. $\endgroup$
    – Onur Oktay
    Commented Jun 11, 2023 at 16:34
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    $\begingroup$ The James space $J$ is complemented in the double dual $J^{**}=J\oplus [1]$ where $[1]=(1,1,1,, \ldots)$. $J$ is a dual space and has explicit description. There is a rich literature on the James space and its variations. Here is a book cambridge.org/core/books/james-forest/… $\endgroup$
    – Bunyamin Sari
    Commented Jun 11, 2023 at 16:55
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    $\begingroup$ We may use Goldstine's theorem as general way of producing other examples. Let $Z$ be any nonreflexive Banach space, $Y=Z^*$, $X=Z^{***}$. It is well-known that $Z^*$ is a closed, complemented, weak$^*$ dense subspace of $Z^{***}$. $\endgroup$
    – Onur Oktay
    Commented Jun 12, 2023 at 11:07
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    $\begingroup$ Perhaps interesting for you is an old observation of Joram Lindenstrauss: If $X$ is complemented in some dual space, then it is complemented in $X^{**}$. $\endgroup$
    – Bill Johnson
    Commented Jun 21, 2023 at 23:40

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