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Let $V$ be an infinite dimensional topological vector space and consider the natural application $\iota\colon V\to V^{**}$. The space $V$ is said to be reflexive if $\iota$ is an isomorphism.

Are there examples where $\iota$ fails to be an isomorphism but $V$ and $V^{**}$ are nevertheless isomorphic?

Can one find an example where $V$ is a Banach space and the isomorphism is actually an isometry?

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  • $\begingroup$ Yes, I remember reading many years ago that there are examples, I think even for the Banach space question. But I. I don't remeber what they are. Sorry I can't be of more help. $\endgroup$
    – Owen Sizemore
    Commented Oct 28, 2010 at 15:50
  • $\begingroup$ So, the inclusion is not an isomorphism (because of lacking surjectivity?), but nevertheles an isomorphism exists? $\endgroup$
    – shuhalo
    Commented Oct 28, 2010 at 16:23
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    $\begingroup$ I'd like to ask a similar question for non-locally compact abelian topological groups and the bi-dual in the sense of character groups: can G and G^^ be isomorphic without the natural map G ---> G^^ being a topological group isomorphism? This is close enough to the question posed that I hope it's okay to ask it here, as the same set of people might know the answer to both. (I read once that nonzero TVS with dual space 0 can be used to create abelian top. groups with trivial char. group, so maybe one can do the same thing for "unnatural" reflexive top. groups.) $\endgroup$
    – KConrad
    Commented Oct 28, 2010 at 16:37
  • $\begingroup$ Exactly Martin. $\endgroup$
    – diverietti
    Commented Oct 29, 2010 at 8:19

2 Answers 2

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Yes, the James space.

This is a good question, and R. C. James is rightly praised for this example.

MR0044024 (13,356d)
James, Robert C.
A non-reflexive Banach space isometric with its second conjugate space.
Proc. Nat. Acad. Sci. U. S. A. 37, (1951). 174–177.

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    $\begingroup$ It would be nice if you could give an idea of how it works here. $\endgroup$
    – diverietti
    Commented Oct 29, 2010 at 8:18
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    $\begingroup$ This construction should be in every textbook on Banach space theory. Why does it need to be added here? jstor.org/pss/2041285 This first page is visible for free, and has a norm for the isomorphic version of the question displayed. The second dual can be described the same way, but change the condition of limit zero to limit exists. Thus the space has codimension 1 in the second dual. $\endgroup$
    – Gerald Edgar
    Commented Oct 29, 2010 at 14:35
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    $\begingroup$ Just because it is not included in every textbook on Banach space theory. Anyway, I just said that it would have been nice to have a brief account here. It doesn't matter! I'll take a look at the paper. Thanks. $\endgroup$
    – diverietti
    Commented Oct 29, 2010 at 15:21
  • $\begingroup$ Might this be in Kalton and Albiac's recent book, perhaps? $\endgroup$
    – Yemon Choi
    Commented Nov 4, 2010 at 6:13
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    $\begingroup$ It is, Yemon, and it is also in Megginson's book and Beauzamy's book 'Introduction to Banach spaces and their geometry'; in the latter case it is covered as a series of exercises. $\endgroup$
    – Philip Brooker
    Commented Nov 16, 2010 at 3:27
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James' construction can be iterated, in order to produce a countable family of pairwise non-isomorphic Banach spaces with the same property.

Look at the following paper:

Marek Wójtowicz - "Finitely Nonreflexive Banach Spaces"

Proceedings of the American Mathematical Society

Vol. 106, No. 4 (Aug., 1989), pp. 961-965.

EDIT. The family $Z_n$ of Banach spaces constructed by Wójtowicz has the following properties:

  1. $Z_n$ is isomorphic to $Z_n^{**}$;
  2. $Z_n$ is $n$-reflexive;
  3. if $n < m$ then $Z_n$ is not isomorphic in $Z_m$ and $1$-complemented in $Z_m$.
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    $\begingroup$ That is not the point of Wojtowicz's paper. You get continuum many examples from James' argument--just change 2 to $p$ for $1<p<\infty$. $\endgroup$
    – Bill Johnson
    Commented Oct 28, 2010 at 16:39
  • $\begingroup$ You are right, I was imprecise since I did not want to write a long answer. Anyway, your comment shows that it is better if I edit it... $\endgroup$
    – Francesco Polizzi
    Commented Oct 28, 2010 at 17:20

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