I do not know how to make this question fully precise (except in ways that make the answer uninteresting), but I have wondered about it for a long time, and it has arisen again in the discussion following my recent post on the Banach space constructed by Spiros Argyros and Richard Haydon. Here I would like to say enough to enable others to think about it. One could think of this post as the beginning of polymath2, but I expect polymath2 to be somewhat different from polymath1 for various reasons: it will involve different people, it may well go much more slowly than polymath1 (I would like to think of it quietly chugging away in the background), and I am not comfortable with some of the mathematics that will be essential to formulating a good conjecture. On a more practical level, I think that having a wiki for polymath 1 (see my blogroll for a link) has worked very well as a way of organizing our collective thoughts on the density Hales-Jewett theorem, and if polymath2 gets off the ground then I would expect an accompanying wiki to be part of it from the very beginning.
A typical argument that shows that a Banach space contains a classical space.
The first thing to say is that all “normal” Banach spaces (by which, for simplicity, I will mean separable Banach spaces of infinite real sequences) contain either or for some For example, a space that is sometimes useful is the Lorentz space where the norm of a sequence is the supremum over all finite sets of . Here is the conjugate index of some (meaning that ) and the norm is designed to agree with the norm on all characteristic functions of finite subsets of (and to be the smallest such norm that has certain symmetry properties).
Here is a sketch of a proof that this Lorentz space contains . You just define a sequence of vectors where each is the characteristic function of a set , multiplied by (the sets are disjoint and this normalization makes into a vector of norm 1), and you do so in such a way that the sizes of the increase very rapidly. Then when you work out the norm of any vector in the subspace generated by the , you have to choose some set and work out . You find that makes a decent-sized contribution to this sum only if has size reasonably comparable to that of , which happens for at most one .
Next, let me give the definition of Tsirelson’s space. Given a vector (that is, infinite sequence) and a finite subset of , we write for the sequence that’s in and 0 outside . In other words, it is the pointwise product of with the characteristic function of . We also write if the maximum element of is less than the minimum element of . Let us call a sequence admissible if and also . Then there are several equivalent ways of defining the norm on Tsirelson’s space. For simplicity we shall define it just for finitely supported vectors (that is, sequences that are eventually zero). To define the whole space one just takes the completion (though that is actually completely unimportant for the problem of whether Tsirelson’s space contains or ).
1. The first definition is that it is the smallest norm such that for every finitely supported vector we have the equation
where the maximum is over all admissible sequences .
2. The second definition is an inductive one. Let us define to be and for each let us define
where again the maximum is over all admissible sequences . It is straightforward to prove inductively that for every , so for each the numbers form an increasing bounded sequence. Therefore they converge to a limit , and it is easy to see that the norm thus defined is the smallest that satisfies the equation given in definition 1.
3. If and are finitely supported non-zero vectors, write if the maximum for which is smaller than the minimum for which . Given a set of finitely supported vectors, let be the set of all vectors of the form such that and is less than or equal to the minimum of the support of . Let be the set of all vectors that are in one coordinate and zero everywhere else. Let be the closure of under the operation . Then
4. Similar to 3 except that instead of talking about the closure of one defines to be and defines to be the union of the
The problem in a nutshell.
Actually, there are two problems, and both have to be solved if there is to be a satisfactory answer to the main question of this post. The first is to find a definition of “explicitly definable” that allows one to define all “normal” spaces such as -spaces, Lorentz spaces, Orlicz spaces, Schreier spaces, etc. (ideally, one could also find an appropriate definition that worked for function spaces as well, so that one could also define spaces, Sobolev spaces, Besov spaces, and so on) but does not allow “inductive” definitions of the Tsirelson type. The second is to show that every space that is explicitly definable according to the given definition contains or .
I have given definition 3 above because I think it has the potential to “combinatorialize” the problem. Note that the coordinates of the vectors in each are all either 0 or negative powers of 2, so they can be identified with infinite sequences of non-negative integers, all but finitely many of which are zero. Thus, one can perhaps focus on weak systems of arithmetic that make it possible to define some sets of such sequences but impossible to define the one naturally associated with Tsirelson’s space (so it would somehow have to outlaw closure operations of the type in 3).
Just to be clear what the combinatorialization is, if we replace each vector in by an integer sequence by replacing 0 with 0 and with , then the operation takes a sequence of already existing finitely supported sequences with the minimum of the support of at least and replaces it by , where is the sequence that adds 1 to all non-zero coordinates. For example, if we already have the sequences and , then we can form the sequence .
Tim Chow is checking with his f.o.m. friends whether anybody has any ideas about this. I look forward to what he has to say when he reports back, or indeed to any direct communication from the f.o.m. people themselves.
At some point I should give the rather nice proof that Tsirelson’s space does not contain or , but strictly speaking it isn’t actually needed for this problem. Perhaps I’ll put it on the wiki when that gets going (if it gets going).