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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.

Martin Sleziak
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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
  • 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 $::m+c:\sim:n+c::$. $;;;;$ –  Feb 25 '12 at 11:20
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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. – Davide Giraudo Feb 25 '12 at 11:25
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    @RickyDemer: Sorry, this is what I meant to write: $n \sim n + k(m-n)$. – Alexander Thumm Feb 25 '12 at 11:32
  • 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. – Davide Giraudo Feb 25 '12 at 11:33
  • If you take $B = \ell^1$, then there is no relation between $0$, $1$ and $2$. I don't know what happens further however.. – Joel Cohen Feb 25 '12 at 11:34
  • @Davide: Do you mean the spaces $B_{2k}$ are pairwise non-isomorphic? $:$ (and the spaces $B_{2k+1}$ are also pairwise non-isomorphic?) $;;$ –  Feb 25 '12 at 11:37
  • @RickyDemer Yes, it's indeed what I mean. – Davide Giraudo Feb 25 '12 at 12:03
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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. – checkmath Feb 25 '12 at 12:46
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    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
  • @t.b. Indeed, what I said is not true. Thanks for the link and the paper. – Davide Giraudo Feb 25 '12 at 22:53
  • 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. – Rabee Tourky Sep 15 '12 at 10:45
  • I remember a lot of discussion about this in the book by Koether on Topological Vector Spaces. – Stephen Montgomery-Smith Dec 27 '12 at 19:00
  • I would try to ask here (maybe they will consider the question decent enough), and hope for an answer of him for complete enlightment. – Julien Mar 20 '13 at 14:25

1 Answers1

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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 $X\cong X''$?

Community
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Sean Eberhard
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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
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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 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 09:50
  • @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.) – Sean Eberhard Mar 21 '13 at 15:27
  • 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). – Philip Brooker Mar 22 '13 at 01:49
  • 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 $m<n$ we have that $X_m$ is a proper subspace of the HI space $X_n$, hence $X_m$ is not isomorphic to $X_n$. – Philip Brooker Mar 22 '13 at 01:56
  • 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.). – Philip Brooker Mar 22 '13 at 02:02
  • Question 2 (at the end of your post) is solved here. – Watson Feb 25 '17 at 21:08
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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. – Sean Eberhard Feb 25 '17 at 22:43