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It's well known that if $X$ is a Banach space, then $X^{**}$ contains a copy of $X$, so we'll informally say $X\subset X^{**}$. It's also well known that in general we don't have $X=X^{**}$. So what if, starting with a Banach space $X$ such that $X\subset X^{**}$ but $X\neq X^{**}$, you considered the sequence $$X\subset X^{**}\subset X^{****}\subset X^{******}\subset\cdots$$ Not sure why this sequence would be of any interest but...is it possible that the inclusions are all strict in this infinite chain?

EDIT: Some are (justifiably) suggesting that this question is a "duplicate" of the question of whether or not $X$ is reflexive iff $X^*$ is reflexive. While, in some sense, one could make that claim based on how the answer to my question almost immediately follows from the truth of that statement, I wouldn't go so far as to claim it's a duplicate question.

Fozz
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    I would hazard this guess. The chain has strict inclusions iff $X$ is not reflexive. – ncmathsadist Jun 04 '17 at 14:39
  • by "has" are you suggesting that "all" the inclusions are strict if $X$ is not reflexive? – Fozz Jun 04 '17 at 14:41
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    Correct. I don't think you can keep double-dualing to get a reflexive space. – ncmathsadist Jun 04 '17 at 14:42
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    Since $X$ is reflexive iff $X^{**}$ is reflexive, if there is equality in one, then there is equality everywhere. – Sahiba Arora Jun 04 '17 at 14:44
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    If we have equality at any point, we have equality everywhere. For $Y$ is reflexive if and only if $Y^{\ast}$ is reflexive. – Daniel Fischer Jun 04 '17 at 14:44
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    Ah i guess the answer is a lot simpler than i thought. From the theorem that daniel mentioned, that $X$ reflexive iff $X^$, an immediate corollary is the result that Sahiba mentioned, that $X$ reflexive iff $X^{*}$ is reflexive...then the claim follows. Someone should go ahead and write it as an answer so i can close this question – Fozz Jun 04 '17 at 15:07
  • Related: https://math.stackexchange.com/questions/113198 – Watson Jun 04 '17 at 16:12
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    An funny example is to start with $c_0(\Bbb N)$. The sequence of duals is: $$c_0(\Bbb N),\ \ell^1(\Bbb N),\ \ell^\infty(\Bbb N),\ C(\beta\Bbb N)$$ $c_0$ and $\ell^1$ are seperable, $\ell^\infty$ isn't and $C(\beta\Bbb N)$ even has a larger cardinality than $\mathfrak c$. This sequence of dual spaces really "explodes" super fast! – s.harp Jun 04 '17 at 17:49

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