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Let $(x_{n})_{n}$ be a normalized basic sequence in $X=L_{p}$, with $1<p<2$.

Does there exist a subsequence $(x_{k_{n}})_{n}$ of $(x_{n})_{n}$ and a weakly null sequence $(x^{*}_{n})_{n}$ in $X^{*}$ such that $(x_{k_{n}})_{n}$ and $(x^{*}_{n})_{n}$ are biorthogonal?

This question may be obvious or stupid. Thank you!

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I think it is worthwhile to point out that you do not need to pass to a subsequence of $(x_n)$. To see that, let $(y_n^*)$ be any Hahn-Banach extensions to $X^*$ of the functionals biorthogonal to $(x_n)$ and observe that all weak$^*$ cluster points of $(y_n^*)$ in $X^*$ are in $(x_n)^\perp$. By the separability of $X$ there is a metric $d$ on $X^*$ that, on bounded subsets of $X^*$, generates the weak$^*$ topology. Let $K= \sup_n \|y_n^*\|$. Then the $d$-distance of $y_n^*$ to the $K$-ball of $X^*$ tends to zero, so you get $u_n^*$ in the $K$-ball of $(x_n)^\perp$ s.t. $d(y_n^*,u_n^*)\to 0$ as $n\to \infty$. Set $x_n^*:= y_n^* -u_n^*$.

I don’t know who first did this (maybe Singer), but this is basically Bill Veech’s argument for Sobczyk’s theorem that $c_0$ is separably injective.

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  • $\begingroup$ I think that you meant $K$-ball in $(x_n)^\perp$. $\endgroup$ – August Cleaner Aug 5 '16 at 5:11
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You can find such subsequence as follows. Denote by $\{f_n\}$ some uniformly bounded extensions of biorthogonal functionals of $\{x_n\}$. This is a bounded sequence in $L_q$, $1/q+1/p=1$. It contains a weakly convergent subsequence $\{f_{k_n}\}$. Let $f$ be its weak limit. Then $\{x_{ k_n}\}$ and $\{f_{k_n}-f\}$ form the desired pair of sequences.

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