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I'm have difficultly nailing down the direction of some implications. For $2 \leq q < \infty$, there are (at least) two ways to say that a Banach space $B$ has "cotype $q$".

  • $B$ has cotype q.
  • $B$ is isomorphic to a $q$-uniformly convex Banach space, i.e. a uniformly convex Banach space with a "power type" modulus of uniform convexity $\delta(\varepsilon) = C \varepsilon^q$. ($B$ is said to have martingale cotype q. The name comes from a characterization by Pisier involving martingales.)

I know the following:

  1. Every $q$-uniformly convex Banach space (and any space isomorphic to it) has cotype $q$.
  2. There are nonreflexive spaces, e.g. $L^1$ and $\ell^1$, with cotype $2$. Since they are nonreflexive they are not isomorphic to a uniformly convex space.
  3. The super-reflexive spaces are exactly those isomorphic to uniformly convex spaces, which in turn are all isomorphic to $q$-uniformly convex spaces.

However, I can't seem to find the answer to the following.

If a space is super-reflexive and of cotype $q$, is it isomorphic to a $q$-uniformly convex space?

If not, is there a nice class of spaces where these two notions of cotype agree?

Update 1: I have some partial answers of spaces for which the notions agree (but not yet a general answer to my question).

  • UMD spaces. (see Cédric's answer).
  • Banach lattices of type $p>1$ (which includes the super-reflexive Banach lattices). I found a pair of interdependence diagrams on pp. 100, 101 of Lindenstrass and Tzafriri's "Classical Banach Spaces II" (these are some of the best math diagrams I have seen). On a Banach lattice, the modulus of convexity is of power type $q$ for an equivalent norm if and only if is of cotype $q$ and there is an upper estimate $p>1$. Following the diagram, "upper estimate" can be replaced with "type".

Update 2: On the bottom of p. 78 of Lindenstrass and Tzafriri's "Classical Banach Spaces II" my main question is listed as an open problem. So I guess my question becomes, has it been solved yet?

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I'll attempt to answer this with what I've found:

(1) The answer to my main question is that it is not true. In this document of Pisier's, he states

It is possible to find a uniformly convex space $B$ for which the index of type $p(B)$ differs from the corresponding index for the $M$-type. Similarly for the cotype. See [P4] for details.

The reference [P4] is

G. PISIER, Un exempte concernant la super-réflexivité. Annexe no. 2. Séminaire Maurey-Schwartz 1974-75. Ecole Polytechnique. Paris.

which I can't seem to find, so I'll have to take his word for it.

(2) As I added to my original post, it seems the two best cases where they agree are UMD spaces (thanks Cédric) and Banach lattices of type $p>1$.

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Let $X$ be a Banach space. Suppose that $X$ has UMD (hence super-refexive). It seems to me that the following equivalence is classical.

Then $X$ is of cotype $q$ if and only if $X$ is $q$-uniformly convex.

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  • $\begingroup$ Thanks. I assume you mean to say "$X$ is isomorphic to a $q$-uniformly convex space". (It seems that $\mathsf{UMD}$ is preserved under isomorphisms, where as uniform convexity is not. That is of course if we are using the same definition of $q$-uniformly convex.) Also, I'm still hoping for an answer to my main question, but maybe $\mathsf{UMD}$ is the best known. $\endgroup$
    – Jason Rute
    May 5, 2013 at 10:50
  • $\begingroup$ Undoubtedly, you must look at the Bourgain's exemple of supereflexive space failing UMD and see... $\endgroup$
    – user33709
    May 5, 2013 at 11:03
  • $\begingroup$ Cédric, I found the article, link.springer.com/article/10.1007%2FBF02384306, but I am not sure I understand your comment. Are you saying if I understood this example, I might be able to answer my question? $\endgroup$
    – Jason Rute
    May 5, 2013 at 20:48
  • $\begingroup$ My feeling is that the answer is false and that you need to study the exemples of super-reflexive spaces which are not UMD. Maybe, one of them is a counter-example to your question. See the recent paper front.math.ucdavis.edu/1112.0739 for information on all known counter-examples. $\endgroup$
    – user33709
    May 6, 2013 at 10:02

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