Timeline for On the non-existence of a boundedly complete basic sequence
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14 at 17:19 | comment | added | S Argyros | In the above the sequence $(y_n)_n$ satisfies the property that there exists $\epsilon > 0$ such that $\epsilon \le \|y_n\| \le 1$. | |
| Oct 14 at 17:07 | comment | added | S Argyros | I think that Gowers Dichotomy yields the following.If 𝑋 is a Banach space with a basis $(𝑒_𝑛)_𝑛$ then either it has a block basic sequence $(𝑥_𝑛)_𝑛$ which is boundedly complete or it has a block basic sequence $(𝑦_𝑛)_𝑛$ such that if $D$ denotes $N$ with the dyadic tree order then for every $s$ finite initial segment of $𝐷$ we have $\|\sum_{a\in s} y_a\| \le 1$. The second alternative reminds the basis of the dual of James tree space. . | |
| Oct 14 at 15:18 | comment | added | S Argyros | @Bill Johnson:I was suspected that there exists an alternative approach of the non separability of $X^* $ and I was expecting something from you. | |
| Oct 14 at 15:04 | comment | added | Bill Johnson | Of course you are right, @SArgyros: $X^*$ must be HI saturated. Also, you are right that "for 𝑋 separable such that 𝑋∗ does not contain boundedly complete basic sequence then 𝑋∗ is non separable.' This is contained in my old paper with Rosenthal, "On weak$^*$ basic sequences...". The proof does not use Ramsey Theory or Gowers Dichotomy. | |
| Oct 14 at 14:09 | comment | added | S Argyros | @C Stuart: This is an interesting question. I believe that it has a negative answer. The candidate space is related to Gowers tree space and it is not explained in a mathoverflow answer.I can give you some more details if you send me an email. | |
| Oct 14 at 13:46 | comment | added | C Stuart | Thank you both for your help. The property I'm trying to use is that either X or X* has a boundedly complete basic sequence, and I was wondering how general it is. I couldn't find any examples that didn't have that property. | |
| Oct 13 at 11:37 | comment | added | S Argyros | I think that for $X$ separable such that $X^*$ does not contain boundedly complete basic sequence then $X^*$ is non separable. In my view this can be proved with the use of Ramsey theory. I mean either the classical infinite Ramsey Theory or Gowers Dichotomy. | |
| Oct 13 at 10:21 | comment | added | S Argyros | @Bill Johnson: Must be HI or it does not contain an unconditional basic sequence? i.e. it is HI saturated? | |
| Oct 12 at 18:29 | comment | added | Bill Johnson | Jog my memory: Is there known to exist a space $X$ such that its dual $X^*$ does not contain a boundedly complete basic sequence? If so, $X^*$ must be HI, but $X^*$ being HI is not sufficient. | |
| S Oct 11 at 21:15 | review | First questions | |||
| Oct 11 at 22:46 | |||||
| S Oct 11 at 21:15 | history | asked | C Stuart | CC BY-SA 4.0 |