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Let $ \big\langle B,+,\cdot, \:\: \|\cdot\| \:\: \big\rangle $ be a Banach space.

Define $ \mathbf{B} \; = \;\big\langle B,+,\cdot, \:\: \|\cdot\| \:\: \big\rangle $.

Define $\: \mathbf{B}_0 = \mathbf{B} \:$. For all non-negative integers $n$,
define $\mathbf{B}_{n+1}$ to be the Banach space that is the continuous dual of $\mathbf{B}_n$.

Define the relation $\:\sim\:$ on $\:\{0,1,2,3,4,5,\ldots\}\:$ by
$m\sim n \:$ if and only if $\: \mathbf{B}_m$ is isometrically isomorphic to $\mathbf{B}_n$.

$\sim\:$ is obviously an equivalence relation.

What can the quotient of $\:\{0,1,2,3,4,5,\ldots\}\:$ by $\:\sim\:$ be?

The only thing I know about this is that $\:\{\{0,1,2,3,4,5,\ldots\}\}\:$
and $\:\{\{0,2,4,6,8,\ldots\},\{1,3,5,7,9,\ldots\}\}\:$ are both possible.

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Let $m > n$ and $n \sim m$, then necessarily $n \sim k(m-n)$ for all $k \in \mathbb N$. – Alexander Thumm Feb 25 '12 at 11:13
It can be shown that either $B$ is reflexive, so $B'$ is also reflexive and we are in the second case, and if $B$ is not reflexive all the sets $B_{2k}$ are distinct. – Davide Giraudo Feb 25 '12 at 11:25
@RickyDemer: Sorry, this is what I meant to write: $n \sim n + k(m-n)$. – Alexander Thumm Feb 25 '12 at 11:32
You have to see that a Banach space may not be reflexive, however it may be isometric to its bidual, Brezis Emphasizes that on his book about functional Analysis. – checkmath Feb 25 '12 at 12:46
A somewhat related thread on MO. @Davide: That's not true. For example, if $X = J \oplus J^\ast$, where $J$ is the James space then $X$ is isomorphic to its dual (you can make it isometric by taking the $2$-norm on the direct sum), while $X$ is not reflexive. (that's probably what chessmath is getting at) – t.b. Feb 25 '12 at 12:57

This is not a complete answer, but rather an attempt to draw attention to the relevant subproblems.

In his comment @AlexanderThumm made the following observation: if $m>n$ and $m\sim n$ then $n+k(m-n)\sim n$ for all nonnegative $k$. This follows from the observation that if $X$ and $Y$ are isometric Banach spaces then so are $X'$ and $Y'$.

It's a simple inference from this that either $\mathbf{B}_m \not\cong \mathbf{B}_n$ whenever $m\neq n$, or there are integers $N$ and $n\leq N$ such that $\mathbf{B}_i$ are pairwise nonisometric for $1\leq i\leq N$ and $\mathbf{B}_{N+1}\cong\mathbf{B}_{n}$, $\mathbf{B}_{N+2}\cong\mathbf{B}_{n+1}$, and so on.

It remains to show that each of these situations is possible, or to rule some of them out.

The first case (all $\mathbf{B}_n$ distinct) can happen, considering the sequence $c_0, \ell^1, \ell^\infty,\ldots$. (It's well known that none of these spaces is reflexive, but we need a better argument to rule out the existence of any isometric isomorphism.) For an arbitrary infinite set $S$ consider the space $\ell^\infty(S)$ of bounded functions on $S$, with the uniform norm. This space has cardinality $2^{|S|}$. It is apparently known however (as cited here) that $\ell^\infty(S)''$ is isomorphic to $\ell^\infty(2^{2^S})$, so $\ell^\infty(S)''$ is not isomorphic to $\ell^\infty(S)$ since $2^{2^{2^{|S|}}}>2^{|S|}$. The same argument shows that the $2k$th dual of $\ell^\infty(S)$ is not isomorphic to $\ell^\infty(S)$. It follows from this that no two of the sequence $c_0,\ell^1,\ell^\infty(\mathbf{N}),\ldots$ are isomorphic: if $\mathbf{B}_m\sim\mathbf{B}_n$, say, then $\mathbf{B}_{2m}\sim\mathbf{B}_{2n}$, so we have a contradiction.

I don't have anything to contribute to the other problems, other than to highlight some interesting first cases:

  1. Is there a Banach space $X$ such that $X\cong X'''$ but $X\not\cong X''$?

  2. Does $X'\cong X'''$ imply $X\cong X''$?

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Nice summary, Sean. I started making notes on the problem when it was first posed, but I didn't write it here due to lack of time and I realised quickly that it is probably a very deep question. My notes say much the same as your answer here with the additional remarks that there are examples of the second type that you mention above for $N-n=1$, e.g., the James tree space $JT$ and its canonical predual $JT_\ast$ are norm separable, but $JT'$ is nonseparable; on the other hand, if $B$ is the $m$th dual of $JT$ for any $m\geq 1$, then $B_m \cong B_{m+2n}$ for any $n\geq 1$. One might be... – Philip Brooker Mar 20 '13 at 23:46
able to get other examples for $N-n=1$, in particular for larger $N$, by iterating the James-Lindenstrauss construction (this may be easy or hard - or impossible! - I haven't tried to check). As for the case $N-n>1$, this is where I think things get to the point where one has to do something awesome to make progress. As pointed out in my comments to Douglas Zare's answer to… , this is related to a special case of the Schroeder-Bernstein problem for Banach spaces that seems to be unsolved – Philip Brooker Mar 20 '13 at 23:50
. Since you seem to be at Cambridge, perhaps you can ask Gowers if he has thought about the $N-n>1$ case and whether $X$ and $X''$ are isomorphic to one another when they are isomorphic to complemented subspaces of one another? I think I mentioned this problem to Bill Johnson on Mathoverflow at some stage and he didn't say that he knew the answer, so I'd say it is probably unknown, or not widely known. – Philip Brooker Mar 20 '13 at 23:55
Sorry, my first comment should say, "if $B$ is the $m$th dual of $JT$ for any $m\geq 1$, then $B\cong B_{2n}$ for any $n\geq 1$". – Philip Brooker Mar 20 '13 at 23:57
@Philip, Thank you for your comments, and your example showing that N=4, n=3 (if I understand correctly) is possible. Very interesting! The real interest seems to be N-n>1, and in particular question 1 that I highlighted. – Sean Eberhard Mar 21 '13 at 9:50

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