4
$\begingroup$

Let $X$ be a separable Banach space and $X^*$ be its dual, let $\{x_i\}$ be a sequence in $X$ with dense linear span and such that there exists a sequence $\{x_i^*\}$ in $X^*$ satisfying $x_i^*(x_j)=\delta_{i,j}$ (Kronecker delta). It is clear that $\{x_i^*\}$ is uniquely determined. Let us call $\{x_i\}$ a Markushevich–Auerbach basis if the linear span of $\{x_i^*\}$ is weak* dense in $X^*$ and $1=\|x_i\|_X=\|x_i^*\|_{X^*}$ for all $i$.

Question: Is the problem about the existence of a Markushevich–Auerbach basis in an arbitrary infinite-dimensional separable Banach space still open?

I am asking this question since I noticed that some recent papers (see Theorem 3.3 in http://arxiv.org/pdf/1604.03547.pdf) claim a solution of this problem, but I do not understand the argument.

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ As far as I know the problem is still open. It is not solved in the linked paper. $\endgroup$
    – Bill Johnson
    Commented Jul 31, 2016 at 18:33
  • 2
    $\begingroup$ 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. $\endgroup$
    – Tomasz Kania
    Commented Jul 31, 2016 at 19:49
  • $\begingroup$ @TomekKania: Would you or Bill Johnson like to post an answer? $\endgroup$
    – Nate Eldredge
    Commented Aug 1, 2016 at 13:38
  • $\begingroup$ @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? $\endgroup$
    – Ben W
    Commented Aug 7, 2016 at 1:58
  • $\begingroup$ @BenWallis I do not understand: why does he author of the linked paper claim that $|\hat x^*_i(y)|\le p_i(y)$? $\endgroup$
    – August Cleaner
    Commented Aug 7, 2016 at 5:20

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .