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Every book which treats dual spaces of normend spaces states that $(c_0)' = \ell^1$ and $(\ell^1)' = \ell^\infty$ and some also describe $(\ell^\infty)'$.

However, is anything known about higher order duals in general? Does taking the normed dual of a given normed space $X$ stabilize? In other words: Defining $X^{(1)} = X'$ and $X^{(n)} = (X^{(n-1)})'$. Is there always an integer $n$ such that $X^{(n+1)} = X^{(n)}$?

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Do you mean X^{(n+2)} = X^{(n)}? If X is reflexive then your condition usually won't hold. –  Qiaochu Yuan Nov 15 '10 at 19:28
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I don't think it can. If it does, $X^{(n)}$ would be reflexive. Since a space is reflexive iff its dual is, this will induct all the way down to $X^{(0)}$. Also, on a intuitive level, you should expect "stabilization" as something with period 2, not with period 1 (think $\ell^p$ and $\ell^q$ with $p\neq 2$ and $q = p'$...) –  Willie Wong Nov 15 '10 at 19:29
    
I don't think the question even makes sense, since the equalities that you list are actually isomorphisms. This distinction causes some subtle problems; unfortunately, I can't think of an especially short and illuminating example right now. –  Thierry Zell Nov 15 '10 at 19:34
    
@Thierry: I guess X can be isometric to X^{(2n)} without the isomorphism being the double dual map applied n times? –  Qiaochu Yuan Nov 15 '10 at 20:45
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@Thierry: I don't think that's a problem, doe to the nice functorial/categorical properties of the dual. I'd suggest that a fully principled way to state it might be something like: is there always some $n$ such that the canonical map $X^{(n)} \to X^{(n+2)}$ is an isomorphism? –  Peter LeFanu Lumsdaine Nov 15 '10 at 21:15
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4 Answers 4

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As noted previously, we can restrict our attention to Banach spaces.
A Banach space $X$ is reflexive if the canonical embedding into its double-dual is onto.

Now let $X$ be any Banach space. We define $X^{(\alpha)}$ for all ordinals $\alpha$ as follows: $X^{(0)}=X$, $X^{(\alpha+1)}=(X^{(\alpha)})^\prime$, and for limit ordinals $\delta$ let $X^{(\delta)}$ be the direct limit of the $X^{(\alpha)}$, $\alpha<\delta$, $\alpha$ is even.

For a given space $\alpha$, we can ask whether $X^{(\alpha)}$ will ever become reflexive.

The key fact is that a closed subspace of a reflexive space is again reflexive. (It was pointed out before that $X$ is reflexive iff its dual is. But for the following discussion we have to think about subspaces of reflexive spaces.) This implies that if $X$ is not reflexive, then the sequence $X^{(\alpha)}$, $\alpha$ ordinal, will never stabilize, i.e., no $X^{(\alpha)}$ will be reflexive, like in the $\ell^\infty$ example above (which can be iterated through all the ordinals).

I find this fact rather striking. It was pointed out to me by either Dirk Werner or Ehrhard Behrends. I don't remember exactly who.

Now remarkably, this needs the axiom of choice (which comes from the use of the Hahn-Banach Theorem in the proof). I have convinced myself a while ago that if you don't have a nontrivial atomless finitely additive probability measure on $\mathbb N$ (which can happen if AC fails), then the sequence for $X=c_0$ is $c_0$, $\ell^1$, $\ell^\infty$, $\ell^1$ and so on.
I.e., $\ell^1$ and $\ell^\infty$ are both reflexive in this situation and duals of each other. This is because without the nontrivial measure, every functional on $\ell^\infty$ that vanishes on $c_0$ vanishes on all of $\ell^\infty$. If I remember correctly, there is some remark about this in Solovay's paper on the model of set theory in which all sets of reals are measurable. In this model there is no nontrivial finite, finitely additive, atomless measure on the integers.

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@Stefan, this is probably obvious, but why is it that at limit stages we cannot have a reflexive space? –  Andres Caicedo Nov 16 '10 at 17:03
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Because the non-reflexive spaces earlier in the iteration are closed subspaces of the limit spaces. And, as I pointed out, every closed subspace of a reflexive space is again reflexive. –  Stefan Geschke Nov 16 '10 at 17:09
    
Sure, of course! Thanks. –  Andres Caicedo Nov 16 '10 at 17:45
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Well, maybe this is barely worth more than a comment. But...

We always have a canonical map $\kappa:X \rightarrow X''$ which is always an isometry onto its range (thanks for Hahn-Banach). Then $X$ is reflexive if $\kappa$ is an isomorphism, that is, $\kappa$ surjects. Any standard book on Banach spaces will prove the slightly non-trivial fact that $X$ is reflexive if and only if $X'$ is.

So, it doesn't really make sense to ask if $X^{(n)} = X^{(n+1)}$. However, we can ask if ever $X^{(n)} = X^{(n+2)}$. If so, then as Willie says, $X^{(n)}$ would be reflexive, and so $X^{(n-1)}$ would be, and so... and so $X$ is reflexive, so actually $X=X^{(2)}$ already.

A famous example of James (http://www.jstor.org/stable/1969430) provides a Banach space $J$ such that $\kappa:J\rightarrow J''$ is not surjective, but we have some other map giving $J\cong J''$ (a slight modification even gives $J$ isometric to $J''$).

Maybe you could ask for an isomorphism $X^{(n)} \cong X^{(n+1)}$. I'm not sure about that...

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There was some discussion of this at mathoverflow.net/questions/43986/… –  Yemon Choi Nov 15 '10 at 22:43
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As noted by Thierry in his comment, the question isn't posed quite as well as it could be. However, the following example shows that taking duals can result in spaces that are topologically larger than earlier duals. Consider the space $\ell_\infty (S)$ of bounded scalar functions on an infinite set $S$, equipped with the supremum norm. Note that $\ell_\infty (S)$ has density $2^{\vert S\vert}$ (with respect to the norm topology), whereas $\ell_\infty (2^{2^{\vert S\vert }})$ has density $2^{2^{2^{\vert S\vert }}}$; in particular, $\ell_\infty (S)$ and $\ell_\infty (2^{2^{\vert S\vert }})$ cannot be isomorphic to one another since $2^{2^{2^{\vert S\vert }}}>2^{\vert S\vert }$. However, H. Rosenthal has shown (Theorem 5.1 of Injective Banach spaces and the spaces $L^\infty (\mu)$ for a finite measure $\mu$, Acta Math 124, 1970, 205--248) that $\ell_\infty(S)^{\ast\ast}$ is isomorphic to $\ell_\infty (2^{2^{\vert S\vert }})$, hence $\ell_\infty(S)$ is not isomorphic to $\ell_\infty(S)^{\ast\ast}$. The point of this example is that it is a counterexample to the OP's question regardless of whether one considers the canonical embedding of $X$ into $X^{\ast\ast}$ or one can consider an arbitrary isomorphism between higher duals.

It is possible, however, to build an entire zoo of spaces that are isometrically isomorphic to all of their dual spaces - just take $X\oplus_2 X^*$, where $X$ is any reflexive space (here the direct sum is with respect to the $\ell_2$ norm on $\mathbb{R}^2$). Perhaps going off on a bit of a tangent, one can generalise this construction by considering the space $Z = \(Y\oplus Y^\ast \oplus Y^{\ast\ast} \oplus Y^{\ast\ast\ast} \oplus \ldots\)_{\ell_2}$ for any Banach space $Y$; if $Y$ is reflexive then $Z$ is isometrically isomorphic to all of its duals. If $Y$ is nonreflexive, then for the $m$th and $n$th duals of $Z$, denoted $Z^{m\ast}$ and $Z^{n\ast}$ respectively (here $m$ and $n$ are nonnegative integers, with $Z^{0\ast}=Z$), we still have that $Z^{m\ast}$ and $Z^{n\ast}$ are isometrically isomorphic to subspaces of one another. Moreover, $Z^{m\ast}$ and $Z^{n\ast}$ are isometrically isomorphic to complemented subspaces of one another if neither $m$ or $n$ is $0$ (or if $Y$ is a dual space). However, it still seems to be an open problem (see Plichko and Wojtowicz's paper Note on a Banach space having equal linear dimension with its second dual, Extract Math. 18(3), 2003, p.311--314, for a statement of this problem in the literature) as to whether $X$ and $X^{\ast\ast}$ being isomorphic to complemented subspaces of one another implies that $X$ and $X^{\ast\ast}$ are isomorphic.

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Phil, I wonder if there is a quasi-reflexive space that is not isomorphic to its second dual. I doubt that this is known or follows from easily from the existing technology. –  Bill Johnson Nov 16 '10 at 14:27
    
Interesting question Bill. Prior to reading your comment I had never wondered about whether quasi-reflexive H.I. spaces exist, but it turns out that they do: Theorem V.1 of the Argyros-Todorcevic book Ramsey Methods in Analysis. Their space $X$ has codimension $1$ in its bidual, but of course the bidual $X^{\ast\ast}$ can't be isomorphic to $X$ (or any of its proper closed subspaces). Thus quasi-reflexive H.I. spaces are examples with the property you mention. Admittedly, I don't know much about H.I. spaces, but luckily I remembered the key fact that they don't embed properly in themselves. –  Philip Brooker Nov 17 '10 at 1:51
    
A related question that comes to mind is whether every quasi-reflexive space is isomorphic to a dual space. –  Philip Brooker Nov 18 '10 at 6:06
    
Actually, every quasi-reflexive space is isomorphic to a dual space of arbitrarily high order; this and many other facts regarding quasi-reflexive spaces are contained in Civin and Yood, Quasi-reflexive spaces, Proc. Amer. Math. Soc. 8 (1957), 906--911. –  Philip Brooker Nov 19 '10 at 6:25
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It does not seem to make much sense to consider normed spaces as opposed to Banach spaces because the dual of a normed space is always complete, so I restrict the discussion to Banach spaces.

Think of the classical $L^{p}$-spaces, where $(L^{p})^{\ast} \cong L^{1}$ with $\frac{1}{p} + \frac{1}{q} = 1$ for $1 < p, q < \infty$. If $1 < p < 2$ and $2 < q < \infty$ then $L^{p}$ and $L^{q}$ are not usually isometrically ismorphic except in very degenerate cases.

It may happen that $X^{(2k-1)} \cong X^{(2k+1)}$ and $X^{(2k)} \cong X^{(2k+2)}$ isometrically (not necessarily via the canonical inclusion in the bidual), as it happens for instance with the James spaces.

The canonical (isometric) inclusion $X^{(n)} \to X^{(n+2)}$ can only be an isomorphism if $X$ is already reflexive because $X$ is reflexive if and only if $X^{\ast}$ is reflexive. Otherwise the Hahn-Banach theorem shows that $X^{(n)}$ is identified a proper subspace of $X^{(n+2)}$ via the canonical inclusion.

Finally, I think that this may well be an open question, because as far as I know no explicit (analytic) identification of the bidual of $L^{\infty}$ is known.

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Did you mean to say "where $(L^p)^* \cong L^q"? –  Mark Bell Nov 15 '10 at 23:05
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