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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.

  • 0
    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

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