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 $E^\*$.



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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.



Is there a topology that makes every basic sequence null?
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.)


