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Let $B$ be a (non-reflexive) Banach space. Denote by $B^{(n)}$ the $n$-fold dual, i.e. $B^{(0)}=B$ and $B^{(n)}=(B^{(n-1)})^\prime$. Does there exist an $n\ge 1$ s.t. $B^{(n)}\cong B$ ? Please explain.

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James' space is an example of non-reflexive Banach space which is isomorphic to it's bidual: James, A non reflexive Banach space isometric with its second conjugate space, Proc. Nat. Acad. Sc. USA, 37 (1951), p.174-177) – Davide Giraudo Oct 11 '12 at 14:51
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This thread (math.stackexchange.com/questions/113198/… this one (mathoverflow.net/questions/46138) can be helpful. – Davide Giraudo Oct 11 '12 at 15:13
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And you can change you nickname, as it's not a stupid question at all. – Davide Giraudo Oct 11 '12 at 15:31
    
Thanks for the link, the second one solves it sufficiently. That's a somewhat surprising answer. I'll keep my nick, in case I ask a stupid question later on ;-). – anon Oct 11 '12 at 15:33
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Note any separable Banach space $X$ with non-separable dual $X^*$ would furnish a counterexample (since a normed space is separable if its dual is). – David Mitra Oct 11 '12 at 15:38
up vote 2 down vote accepted

It may happen that $B^{(2)}\cong B$, even if $B$ is not reflexive, as James' space shows, see James, A non reflexive Banach space isometric with its second conjugate space, Proc. Nat. Acad. Sc. USA, 37 (1951), p.174-177) for instance.

But it's possible such a $n$ doesn't exists.

Lemma: If $E$ is a Banach space and $E'$ its topological dual space, and $E'$ is separable, so is $E$.

If $\{f_n\}$ is dense in the unit ball of $E'$, for each $n$, fix $x_n\in E$ of norm $1$ such that $|f_n(x_n)| \geq \frac 12\lVert f_n\rVert_{B'}$. We have to show that $F:=\operatorname{Span}\{x_k,k\geq 1\}$ is dense in $E$. Let $f$ a continuous linear functional with vanishes on $F$. We have to show that it vanished on $E$. Let $\{n_k\}$ such that $\lVert f-f_{n_k}\rVert\leq k^{-1}$ (we can assume that $f$ has a norm $\leq 1$). Then $$\lVert f_{n_k}\rVert\leq \lVert f_{n_k}(x_{n_k})-f(x_{n_k})\rVert\leq k^{-1},$$ so $\lVert f\rVert\leq 2k^{-1}$.

To apply this to our problem, take $B$ a separable Banach space with non-separable dual (as $\ell^1(\Bbb R)$). It's not reflexive, otherwise $B''$ would be separable and so would be $B'$. If $B^{(n)}\cong B$ for some $n\geq 2$, then $B^{(n)}$ would be separable, as this property is conserved by isomorphism. By the lemma $B^{(n-1)}$ would be separable, and iterating that $B'$^would be separable, a contradiction.

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Right, fixed now. Thanks. – Davide Giraudo Oct 12 '12 at 18:26

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