Timeline for Does every separable Banach space have a Markushevich–Auerbach basis?
Current License: CC BY-SA 3.0
11 events
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Aug 8, 2016 at 16:07 | comment | added | Bill Johnson | The ``proof" in the linked paper is not a proof, Ben. | |
Aug 7, 2016 at 5:20 | comment | added | August Cleaner | @BenWallis I do not understand: why does he author of the linked paper claim that $|\hat x^*_i(y)|\le p_i(y)$? | |
Aug 7, 2016 at 1:58 | comment | added | Ben W | @BillJohnson I am not sure I understand. In Theorem 3.3 of the linked paper, we get a Markushevich basis $(x_n,x_n^*)_{n=1}^\infty$ satisfying $\|x_n\|\cdot\|x_n^*\|=1$. If we write $y_n=\frac{x_n}{\|x_n\|}$ and $y_n^*=\|x_n\|x_n^*$ then $(y_n,y_n^*)_{n=1}^\infty$ is a Markushevich basis satisfying $\|y_n\|=\|y_n^*\|=1$. Doesn't this solve the problem? If not, where have I gone wrong? Or are you instead saying that the proof of Theorem 3.3 is invalid? | |
S Aug 1, 2016 at 13:41 | history | suggested | Martin Sleziak |
added top-level tag; http://meta.mathoverflow.net/questions/1457/why-are-mo-tags-formatted-as-they-are
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Aug 1, 2016 at 13:38 | comment | added | Nate Eldredge | @TomekKania: Would you or Bill Johnson like to post an answer? | |
Aug 1, 2016 at 13:25 | review | Suggested edits | |||
S Aug 1, 2016 at 13:41 | |||||
Aug 1, 2016 at 12:08 | history | edited | Tomasz Kania | CC BY-SA 3.0 |
edited body; edited tags; edited title
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Jul 31, 2016 at 19:49 | comment | added | Tomasz Kania | Yes, it is still open. Btw. Theorem 2.1 is an easy exercise (certainly known) and the claim that Grothendieck [GR] proved that if a Banach space had the approximation property, then it would also have a S-basis is false. | |
Jul 31, 2016 at 18:33 | comment | added | Bill Johnson | As far as I know the problem is still open. It is not solved in the linked paper. | |
Jul 31, 2016 at 18:11 | history | edited | August Cleaner | CC BY-SA 3.0 |
deleted 78 characters in body
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Jul 31, 2016 at 17:28 | history | asked | August Cleaner | CC BY-SA 3.0 |