Is there a topology that makes every basic sequence null?

Question

Let $E$ be a Banach space. Let $F$ be the collection of all $f\in E^*$ such that $\left<f,e_n\right>\to 0$, for every normalized basic sequence $\{e_n\}$. It is easy to see that $F$ is a closed subspace of $E^*$.

Does $F$ separate points of $E$?

Note that if $E$ is reflexive, then $F=E^*$, since every basic sequence is shrinking (which seems too strong).

In general this is not the case: if $E=l_{\infty}(\mathbb{Z})$, then $F\cap l_1=\{0\}$. Let $f=(f_n)\in l_1$. WLOG an infinite number of $f_n$ nonnegative (otherwise replace $f$ with $-f$). By rearranging the coordinates, we may assume that $f_n\ge 0$, when $n>0$.

Take the Rademacher sequence $r_1=(...,0,0,1,-1,1,-1,...)$, $r_2=(...,0,0,1,1,-1,-1,1,1,...)$, $r_3=(...,0,0,1,1,1,1,-1,-1,-1,-1,...)$ and so on, which is a basic sequence in $l_\infty$.

Let $n$ be such that $f_1+...+f_n>\frac{2}{3}\sum_{n=1}^{\infty}f_n$. Then, for any $m$ such that $2^m\ge n$, $\left<f,r_m\right>=f_1+...+f_n\pm f_{n+1}\pm f_{n+2}...>\frac{1}{3} \sum_{n=1}^{\infty}f_n\not\to 0$.

If $E=C[0,1]$, taking variations of Schauder's basis shows that $F$ does not contain neither discrete measures, nor the Lebesgue measure.

Answer 1

The answer is negative in every non reflexive space. If $X$ is non reflexive, there is a normalized basic sequence $(z_n)$ in X s.t. $(z_1 - z_n)_{n=2}^\infty$ and $(z_1 + z_n)_{n=2}^\infty$ are both basic sequence (necessarily semi normalized). If $x^*$ tends to zero along both of these basic sequences, then $\langle x^*, z_1\rangle =0$.

Actually, $z_1$ can be any unit vector, so $F=\{0\}$. Take a non reflexive subspace $E_1$ of $E$ that does not contain $z_1$ and let $(z_n)_{n=2}^\infty$ be a normalized type $\ell^+$ basic sequence in $E_1$.

(A basic sequence $(y_n)$ is type $\ell^+$ provided there is a constant $\delta>0$ s.t. whenever $(a_n)$ is sequence of non negative scalars, only finitely many of which are not zero, then $$ \| \sum a_ny_n\| \ge \delta \sum |a_n|. $$ Non reflexivity is equivalent to containing a normalized type $\ell^+$ basic sequence.)

Answer 2

This answer is supplementary to the one of Bill Johnson, to fill in some details.

A sequence $\{e_n\}$ in a Banach space $E$ is called a basic sequence of type P* if (among other equivalent definitions) $0<\inf\|e_n\|\le\sup\|e_n\|<+\infty$ and there is $r>0$ such that for any $a_1,...,a_n\in\mathbb{R}$ we have $$|a_1+...+a_n|\le r\|a_1e_1+...+a_ne_n \|.$$

In the paper Singer - Basic sequences and reflexivity of Banach spaces the author showed that reflexivity is equivalent to non-existence of P* basic sequences.

Let $E$ be non-reflexive, let $f\in E^*$ and let $e\in E$ be such that $f(e)\ne 0$. Since the kernel of $f$ is not reflexive there is a P* basic sequence $\{e_n\}$. Since the distance from $e$ to the kernel of $f$ is positive, both of the sequences $\{e-e_n\}$ and $\{e+e_n\}$ are bounded from above and from below.

As $f(e)\ne 0$, at least one of the sequences $\left<f,e+e_n\right>$ and $\left<f,e-e_n\right>$ does not converge to $0$. Hence, it is left to show that $\{e-e_n\}$ and $\{e+e_n\}$ are basic. In order to do that we need to show that there is $K>0$ such that for every $a_1,...,a_n\in\mathbb{R}$ and $m\le n$ we have $$\|a_1(e+e_1)+...+a_m(e+e_m)\|\le K\|a_1(e+e_1)+...+a_n(e+e_n) \|.$$ Indeed, WLOG the norm of $E$ is $l_1$ sum of $e$ and $Ker(f)$, from where

$$\|a_1(e+e_1)+...+a_m(e+e_m)\|=|a_1+...+a_m|+\|a_1e_1+...+a_me_m\|\le $$ $$\le(r+1)\|a_1e_1+...+a_me_m\|\le (r+1)L\|a_1e_1+...+a_ne_n\|\le $$ $$\le(r+1)L(\|a_1e_1+...+a_ne_n\|+|a_1+...+a_m|)=K\|a_1(e+e_1)+...+a_n(e+e_n)\|,$$

where $K=(r+1)L$, and $L$ is the basis constant of $\{e_n\}$.