Let $E$ be an infinite dimensional Banach space, let $E^{\ast}$ denote its continuous (i.e., Banach space) dual, and let $E'$ be its algebraic dual. Clearly, $E^{\ast}$ is a proper vector subspace of $E'$. Now, let us suppose that $E^{\ast}$ 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.

  • $\begingroup$ (If the subspace is topologically isomorphic, it will be algebraically isomorphic, no?) $\endgroup$ Feb 1, 2010 at 22:07
  • $\begingroup$ Well, that's the definition, isn't it ? When you are saying that two normed spaces are isomorphic, are you omitting the linearity ? :-) $\endgroup$
    – Ady
    Feb 1, 2010 at 22:20
  • $\begingroup$ 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? $\endgroup$ Feb 1, 2010 at 22:22
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    $\begingroup$ 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? $\endgroup$ Feb 1, 2010 at 22:25
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    $\begingroup$ 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 :-) $\endgroup$ Feb 2, 2010 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^* $.

  • $\begingroup$ $\kappa^+$ stands for what again? Successor? $\endgroup$ Feb 4, 2010 at 12:14
  • $\begingroup$ $kappa^+$ is the successor of $\kappa$; it is standard set-theoretic notation. Sorry for not explaining in the main text. $\endgroup$ Feb 4, 2010 at 12:18
  • $\begingroup$ Very nice, Pandelis! I'm making some editing (in order to make your life harder ;-)). $\endgroup$
    – Ady
    Feb 5, 2010 at 1:43
  • $\begingroup$ Nice. The argument is a very complicated variation of the example at mathoverflow.net/questions/10993/… :) $\endgroup$ Feb 5, 2010 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 $\ell_2(\gamma)$.

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    $\begingroup$ 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. $\endgroup$
    – Ady
    Feb 3, 2010 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.

  • $\begingroup$ (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.) $\endgroup$ Feb 4, 2010 at 3:20
  • $\begingroup$ Oh, in case it is confusing, the examples Ady gave are now in the comment by Scott Morrison, $l^\infty(\mathbb{N})$ and $C[0,1]^*$. $\endgroup$ Feb 4, 2010 at 3:36
  • $\begingroup$ 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 $\endgroup$
    – Yemon Choi
    Feb 4, 2010 at 8:30

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