most recent 30 from http://mathoverflow.net 2013-05-23T22:25:55Z http://mathoverflow.net/feeds/user/7872 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69542/uniformly-convex-spaces/71053#71053 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>

http://mathoverflow.net/questions/34087/selecting-basic-sequences/35023#35023 Adi Tcaciuc 2010-08-09T17:07:27Z 2010-08-09T17:07:27Z

<p>Indeed, such sequences exist. Sequences that have the same closed span as any of their subsequences appear in the literature under the name <em>overcomplete</em> or <em>overfilling</em> (see for example <em>Byorthogonal Systems in Banach Spaces</em>, by Hajek, Montesinos, Vanderwerff and Zizler, Excercise 1.1, page 42). Every infinite dimensional separable Banach space contains such a sequence whose span is also dense. If $(x_n)_n$ is a set on the unit sphere whose span is dense in $X$, take, for example, $y_n=\sum_{k=0}^{\infty}\frac{x_k}{n^{k}k!}$. Then $(y_n)_n$ converges to $x_0$. Setting $z_n=n(y_n-x_0)$, we see that $(z_n)_n$ converges to $x_1$, and so on...Hence, any subsequence of $(y_n)_n$ has dense span. </p> <p>Taking any normalized, $\omega$-linearly independent, overcomplete sequence and doing the previous construction on the binary tree suggested by Bill, indeed gives a linearly independent uncountable set that contains no infinite minimal subset. </p> <p>Bill's answer was very helpful and completely answers my question. Thanks again. </p>

http://mathoverflow.net/questions/34087/selecting-basic-sequences Adi Tcaciuc 2010-08-01T05:10:13Z 2010-08-09T17:07:27Z

<p>Suppose $(x_\alpha)_\alpha$ is an uncountable, linearly independent family of norm one vectors in a Banach space. Can one always select a basic sequence (or at least a minimal system) from this family? I suspect the answer is no but I cannot come up with an example.</p> <p>Thank you!</p>

http://mathoverflow.net/questions/96832/almost-isometric-subspaces-of-ell-p/96847#96847 Adi Tcaciuc 2012-05-13T20:00:42Z 2012-05-13T20:00:42Z

Yes, and the question as posed, has nothing to do with $l_p$. It is just rescaling an isomorphism $T$ such that $||T||$ is small, and consequently $||T^{-1}||$ is large. I suspect that he meant $1/(1+\epsilon)$ instead of $1-\epsilon, in which case <i>I think</i> it is not true if $S$ is required to be an isomorphism between $X$ and $Y$, but it is true for all $\epsilon&gt;0$ if is not.

http://mathoverflow.net/questions/96044/strongly-convergence-in-reflecxive-banach-space/96051#96051 Adi Tcaciuc 2012-05-09T17:49:29Z 2012-05-09T17:49:29Z

Just to add to Nik answer, reflexivity is invariant to equivalent renormings (topologic property), while uniform convexity is not, it depends on the geometry of the norm (geometric property). The hypotheses that $||f_n||\to||f||$ is also a geometric property, and the conclusion that $f_n\to f$ is again topologic. This should give you the intuition, before any concrete counter-example, that the property shouldn't hold under the assumption of reflexivity, or any other topologic property for that matter.