My first question here would fall into the 'ask Johnson' category if there was one (no pressure Bill). I'm interested in constructing a uniformly convex Banach space with conditional structure without using interpolation. The constructions of Ferenczi and Maurey-Rosenthal both use interpolation.

Using existing methods for constructing spaces with conditional structure I think it is possible to construct a hereditarily indecomposable space whose natural basis statisfies a lower $\ell_2$ estimate on any $n$ disjointly supported blocks vectors supported after the $n^{th}$ position on the basis and an upper $\ell_2$ estimate on all finite block sequences. The space $X$ is sure to be reflexive and probably doesn't contain $\ell_\infty$ finitely represented.

I would like to have some way of showing that $X$ is uniformly convex and this is where I'm stuck. Perhaps one could show that $\ell_1$ is not finitely represented in $X$ but as far as I can see this is not good enough (or is it?).

My question: If a space is reflexive and does not contain $\ell_1$ finitely represented is it necessarily uniformly convex?

I suspect the answer is no but I don't have a counterexample.

Another question: Are there any known conditions on a basis, which (1) do not imply the basis is unconditional and (2) do imply the space is uniformly convex?

  • $\begingroup$ The question needs reformulating slightly, since you can just add a 2D space that's not even strictly convex and it won't change whether $\ell_1$ is finitely represented. $\endgroup$
    – gowers
    Jul 5, 2011 at 16:20
  • $\begingroup$ Oh, and it also won't change whether it's reflexive. So the answer to your question as asked is trivially no. Do you mean "isomorphic to a space that is uniformly convex"? $\endgroup$
    – gowers
    Jul 5, 2011 at 16:21
  • 4
    $\begingroup$ Best tag edit ever. $\endgroup$ Jul 6, 2011 at 5:44
  • 4
    $\begingroup$ What about a 'no pressure Bill' tag? $\endgroup$ Jul 6, 2011 at 5:45
  • 4
    $\begingroup$ Bill's bill is in the mail. :) $\endgroup$ Jul 6, 2011 at 13:34

2 Answers 2


Kevin, there are non reflexive spaces with non trivial type--even of type 2. James constructed the first one; his argument is very complicated. Later Pisier-Xu did it much more simply using interpolation between $\ell_1$ and $\ell_\infty$, but using the universal non weakly compact operator instead of the formal identity between the two spaces. See

Random series in the real interpolation spaces between the spaces vp. Geometrical aspects of functional analysis (1985/86), 185–209, Lecture Notes in Math., 1267, Springer, Berlin, 1987.

For a reflexive space with non-trivial type that is not superreflexive take the $\ell_2$ sum of all finite dimensional subspaces of the Pisier-Xu space.

  • $\begingroup$ Incidentally, the Pisier-Xu space is a prime candidate for a non reflexive space all of whose subspaces have the approximation property. $\endgroup$ Jul 5, 2011 at 18:51
  • $\begingroup$ Do you mean non-reflexive spaces with non-trivial type? Is there is a reflexive space with non-trivial type that is not superreflexive? $\endgroup$ Jul 5, 2011 at 20:14
  • $\begingroup$ Sorry; non reflexive, of course. I'll edit my answer. BTW: Are you coming to our Workshop this year? $\endgroup$ Jul 5, 2011 at 21:04
  • $\begingroup$ Sorry; I wrote my first answer in a hurry and have edited it to explain the relevance of the James/Pisier-Xu work to your question. $\endgroup$ Jul 5, 2011 at 21:10

I think James also showed that if $X$ does not contain almost isometric copies of $\ell_1^2$ (he called such a space uniformly non-square) then $X$ is superreflexive. This is no longer true for $n>2$, as James later constructed a non-reflexive, uniformly non-octahedral (no almost isometric copies of $\ell_1^3$) space, thus also having non-trivial type.

Maybe you can check whether your space is uniformly non-square. Connecting it with your last question I think that you would have to verify that $\exists \delta>0$ such that for any normalized block vectors $x$ and $y$ (but not necessarily disjointly supported) there exist a choice of signs such that $||x\pm y||<2-\delta.$ I don't think this condition implies unconditionality.

Hopefully this makes sense...


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