Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $E$ be an infinite dimensional Banach space, let $E^{\*}$ denote its continuous (i.e., Banach space) dual, and let $E^'$ be its algebraic dual. Clearly, $E^{\*}$ is a proper vector subspace of $E'$. Now, let us suppose that $E^{\*}$ and $E'$ are algebraically isomorphic (i.e., as vector spaces). Does it follow that $E$ contains an isomorph of the Banach space $\ell_{1}(\mathbb{R})$ ?

[By "*isomorph of X" I mean a closed linear subspace both algebraically and topologically isomorphic to X.]

P.S. This is under ZFC + CH.

P.P.S. The answer is affirmative if $E$ is the dual of a separable [infinite-dimensional] Banach space. It would be interesting to see if it is also affirmative when $E$ is a "nice" space. For instant, a Banach lattice.

share|improve this question
    
(If the subspace is topologically isomorphic, it will be algebraically isomorphic, no?) –  Mariano Suárez-Alvarez Feb 1 '10 at 22:07
    
Well, that's the definition, isn't it ? When you are saying that two normed spaces are isomorphic, are you omitting the linearity ? :-) –  Ady Feb 1 '10 at 22:20
    
I'm just wondering why you phrased the definition ion your last sentence in the way you did: how could the subspace be topologically isomorphic and not algebraically isomorphic? –  Mariano Suárez-Alvarez Feb 1 '10 at 22:22
2  
I'm no Banachist, but the hypothesis that the algebraic and topological duals have the same dimension seems very unlikely to me. (At the very least, this cannot happen for a reflexive space.) Do you know an example of an infinite-dimensional Banach space with this property? –  Pete L. Clark Feb 1 '10 at 22:25
2  
Ady had replied, in part, "@Yemon E = the space of all bounded sequences. Or, E = the dual of C[0,1]." I've taken the liberty of deleting the rest of the comment, and the ensuing conversation. Please complain on meta :-) –  Scott Morrison Feb 2 '10 at 2:32

3 Answers 3

Ok Ady, since you like CH I will work with CH, and to make your life easier, I will work with GCH.

Since I do not expect that everybody in MO is aware of various Banach space constructions, let me give some information on James tree spaces which are relevant to the question.

A tree is a partially order set $(T,<)$ such that for every $t$ in $T$ the initial segment $\{s\in T: s < t\}$ is well-ordered under $ < $. A segment of $T$ is a subset $S$ of $T$ which is:

  1. linearly ordered under $ < $ and
  2. for all $s, t, w\in T$ if $s < t < w$ and $s, w \in S$ then $t\in S$.

The completion of $T$, usually denoted by $c(T)$, is the collection of all initial segments of $T$ ordered by inclusion. Notice that $c(T)$ contains $T$ and is much larger than $T$. For instance, if $T$ is the tree of all finite sequences of natural numbers (usually called the Baire tree, which is clearly countable), then its completion is the Baire-tree together with its branches (i.e. the Baire space) and so it has the cardinality of the continuum.

For every tree $T$ the corresponding James tree space $JT$ is defined to be the completion of $c_{00}(T)$ with the norm: $$\|v\| = \sup\{ (\sum_{i=1}^d (\sum_{t\in S_i} v(t) )^2 )^{1/2} \}$$ where the above supremum is taken over all finite families $(S_i)_{i=1}^d$ of pairwise disjoint segments of $T$. Basic facts (I can provide appropriate references to anyone who is interested):

  • For every tree $T$ the space $JT$ is hereditarily $\ell_2$; that is, every infinite-dimensional subspace of $JT$ contains a copy of $\ell_2$.
  • For every tree $T$ the second dual of $JT$ is linearly isometric to the James tree space of the completion $c(T)$ of $T$. In particular, neither $JT^* $ nor $JT^{**}$ contain a copy of $\ell_1$.

Now we come to the specifics of the construction. Remember that we work with GCH. This implies, in particular, the following: if $X$ is a Banach space of cardinality $\kappa$, then the algebraic dual of $X$ has cardinality $\kappa^+$.

Let $T$ be the tree of all countable subsets of $\omega_1$ equipped with the partial order of end-extension. We have GCH, hence, the tree is just all sequences of real numbers, and so, it has cardinality $\aleph_1$. The cardinality of the corresponding James tree space is also $\aleph_1$.

The completion $c(T)$ of our tree $T$ is the set of all subsets of $\omega_1$. Hence it has cardinality $2^{\aleph_1}$ which is, under GCH, $\aleph_2$. It follows that the cardinality of $JT^{**}$ is $\aleph_2$.

Now consider cases.

Case 1: the topological dual $JT^* $ of $JT$ has cardinality strictly bigger than $\aleph_1$. Then we are done: our counterexample is $JT$.

Case 2: the topological dual $JT^* $ of $JT$ has cardinality $\aleph_1$. We are also done: our counterexample is $JT^* $.

share|improve this answer
    
$\kappa^+$ stands for what again? Successor? –  Harry Gindi Feb 4 '10 at 12:14
    
$kappa^+$ is the successor of $\kappa$; it is standard set-theoretic notation. Sorry for not explaining in the main text. –  Pandelis Dodos Feb 4 '10 at 12:18
    
Very nice, Pandelis! I'm making some editing (in order to make your life harder ;-)). –  Ady Feb 5 '10 at 1:43
    
Nice. The argument is a very complicated variation of the example at mathoverflow.net/questions/10993/… :) –  Mariano Suárez-Alvarez Feb 5 '10 at 1:59

Maybe we can look at $\ell_2(\gamma)$, where $\gamma$ is an uncountable set. The topological dual is itself. Its algebraic dual seems to have the same cardinality as $\ell_2(\gamma)$.

share|improve this answer
3  
It is not hard to see that $\left|\ell_{2}\left(\gamma\right)\right|=\left|\gamma\right|$ in this case. OTOH, any mapping from $\gamma$ to $\mathbb{R}$ can be "extended'' to a linear (discontinuous) functional, so that $2^{\left|\gamma\right|}$ would be the dimension of the algebraic dual. –  Ady Feb 3 '10 at 23:49

I think that the cardinality of E' should always be greater than the cardinality of $E^*$, so they never will be isomorphic in any sense. Basically, as pointed out here, $E^*$ is the space of all maps from a topological basis of E into a field. E' is analogously the space of all maps from an algebraic basis to a field. So, this boils down to the question:

In an infinite-dimensional Banach space, does an algebraic basis ever have the same cardinality as a topological basis?

I think the answer is no. For example, in $l^2$ (which is the smallest infinite-dimensional Banach space), the topological basis is countable. As Ady points out, the algebraic basis should have cardinality $2^{|\mathbb N|}$.

I'll community wiki this because this is a guess rather than a proof. Feel free to edit & improve. I seem to remember I knew a slick proof of Ady's statement; if I actually remember it, I'll put it in.

share|improve this answer
    
(1)What about the examples Ady gives in the comments above? (2)What in general do you mean by topological basis? For Hilbert space I assume you mean orthonormal basis, and more generally some Banach spaces have Schauder bases, but I don't know if an analogous set exists in every Banach space. (3) Your characterization of the topological dual isn't accurate; you can't send the "topological basis" just anywhere. E.g., consider the identification of the dual of $l^2$ with $l^2$. (This point, however, only shows that the topological dual is "smaller" than you indicated.) –  Jonas Meyer Feb 4 '10 at 3:20
    
Oh, in case it is confusing, the examples Ady gave are now in the comment by Scott Morrison, $l^\infty(\mathbb{N})$ and $C[0,1]^*$. –  Jonas Meyer Feb 4 '10 at 3:36
    
FWIW not every Banach space has a Schauder basis. If memory serves rightly, the first counter-example was due to Per Enflo. That said, the gap between Schauder basis and Hamel basis is vast in infinite dimensions –  Yemon Choi Feb 4 '10 at 8:30
    
Since it is community wiki and wrong, I voted it down, I hope you understand. If you do remember the slick proof, I'll be happy to undo the vote. –  Harry Gindi Feb 4 '10 at 12:12
    
Since -1 is enough, I voted it up (after previously abstaining). –  Jonas Meyer Feb 6 '10 at 23:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.