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(ZFC)


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 $k \in \mathbb N$.2012-02-25
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    How is that shown? $\;\;\;$ However it is, that makes me notice the obvious constraint that for $\hspace{.8 in}$ non-negative integers $m,n,c$, if $\: m\sim n \:$ then $\:\:m+c\:\sim\:n+c\:\:$. $\;\;\;\;$2012-02-25
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    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.2012-02-25
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    @RickyDemer: Sorry, this is what I meant to write: $n \sim n + k(m-n)$.2012-02-25
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    I forget to add that in the case $B$ non reflexive, $B'$ is non reflexive and so the sets $B_{2k+1}$ are not pairwise isomorphic.2012-02-25
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    If you take $B = \ell^1$, then there is no relation between $0$, $1$ and $2$. I don't know what happens further however..2012-02-25
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    @Davide: Do you mean the _spaces_ $B_{2k}$ are _pairwise_ _non-isomorphic_? $\:$ (and the spaces $B_{2k+1}$ are also pairwise non-isomorphic?) $\;\;$2012-02-25
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    @RickyDemer Yes, it's indeed what I mean.2012-02-25
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    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.2012-02-25
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    A somewhat related [thread on MO](http://mathoverflow.net/questions/46138/). @Davide: That's not true. For example, if $X = J \oplus J^\ast$, where $J$ is the [James space](http://www.pnas.org/content/37/3/174.full.pdf+html) 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)2012-02-25
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    @t.b. Indeed, what I said is not true. Thanks for the link and the paper.2012-02-25
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    Banach spaces with pre-duals are interesting for this problem and I don't think can be avoided here. Some Banach spaces have preduals that are not isometric to each other.2012-09-15
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    I remember a lot of discussion about this in the book by Koether on Topological Vector Spaces.2012-12-27
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    I would try to ask [here](http://mathoverflow.net/) (maybe they will consider the question decent enough), and hope for an answer of [him](http://mathoverflow.net/users/2554/bill-johnson) for complete enlightment.2013-03-20

1 Answers 1

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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...2013-03-20
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    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 http://mathoverflow.net/questions/99777/does-x-embed-in-y-and-y-embed-in-x-always-imply-that-x-isomorphic-on , this is related to a special case of the Schroeder-Bernstein problem for Banach spaces that seems to be unsolved2013-03-20
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    . 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.2013-03-20
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    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 $n\geq 1$".2013-03-20
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    @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.2013-03-21
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    @PhilipBrooker Possibly an easier question than 1 is the following. Is there a Banach space $X$ such that $X''''\cong X$ but $X''\not\cong X$? Gowers suggested an idea for an example, but which requires a bi-infinite sequence $(X_n)_{n\in\mathbf{Z}}$ of Banach spaces such that $X_{n+1}=X_n''$ for all $n$ and $X_m \not\cong X_n$ for all $m\neq n$. Is there an example of such a sequence? (The idea is then to look at a sum of every other term, and hopefully to prove that it is distinct from its double dual.)2013-03-21
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    If one relaxes the condition that $X_{n+1}$ be isometrically isomorphic to $X_n^{''}$ to the condition that $X_{n+1}$ be isomorphic to $X_n^{''}$, then such a bi-infinite sequence certainly exists: first take $X_0$ to be a quasi-reflexive $HI$ space, such as that constructed in Chapter V of the first part of the Argyros-Todorcevic book *Ramsey Methods in Analysis*. For $n\geq 1$ we take $X_n$ to be the $2n$th dual of $X_0$. If $n<0$ and we have constructed $X_{n+1}$, then for $X_n$ take an *isomorphic* second-predual of $X_{n+1}$ (see Civin and Yood's 1957 PAMS paper *Quasi-reflexive spaces*).2013-03-22
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    Notice that each $X_n$ is quasi-reflexive (see also Theorem 3.5 of the Civin-Yood paper) and HI: for $n<0$ it is trivial that $X_n$ is HI, whilst for $n>0$ it follows from the fact that $X_n$ is a compact (hence strictly singular) extension of $X_0$. Finally, notice that for integers $m$X_m$ is a proper subspace of the HI space $X_n$, hence $X_m$ is not isomorphic to $X_n$. – 2013-03-22
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    Regarding the *isometric* property, I do not know off the top of my head whether the Argyros example admits an isometric $2n$th predual for all $n$, but I might have a closer look at it later to see if it is so. The fact that every non-reflexive Banach space can be renormed to be not isometrically isomorphic to the dual of any Banach space (a result of Bill Davis and Bill Johnson) means that there is definitely something to check here (if having the isometric property hold actually matters for Gowers' idea.).2013-03-22
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    Question 2 (at the end of your post) is solved [here](http://math.stackexchange.com/questions/152343).2017-02-25
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    @Watson The question here is a bit different. When here we write $X''\cong X$, we do not necessarily mean via the canonical map, so this is not the same as reflexivity.2017-02-25