Uniformly Convex spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:47:28Z http://mathoverflow.net/feeds/question/69542 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69542/uniformly-convex-spaces Uniformly Convex spaces Kevin Beanland 2011-07-05T15:55:08Z 2011-07-23T06:31:03Z <p>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. </p> <p>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. </p> <p>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?). </p> <p>My question: If a space is reflexive and does not contain $\ell_1$ finitely represented is it necessarily uniformly convex? </p> <p>I suspect the answer is no but I don't have a counterexample. </p> <p>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? </p> http://mathoverflow.net/questions/69542/uniformly-convex-spaces/69559#69559 Answer by Bill Johnson for Uniformly Convex spaces Bill Johnson 2011-07-05T18:14:20Z 2011-07-05T21:05:00Z <p>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</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/69542/uniformly-convex-spaces/71053#71053 Answer by Adi Tcaciuc for Uniformly Convex spaces Adi Tcaciuc 2011-07-23T06:31:03Z 2011-07-23T06:31:03Z <p>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 $X$ <strong>is</strong> 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. </p> <p>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 $||x\pm y||&lt;2-\delta.$ I don't think this condition implies unconditionality. </p> <p>Hopefully this makes sense...</p>