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 $E=l_{\infty}$, then $F$ contains no positive elements of $l_1$. Let $f=(f_1,f_2,...)\in l_1$ be positive, and take the Rademacher sequence $r_1=(1,-1,1,-1,...)$, $r_2=(1,1,-1,-1,1,1,...)$, $r_3=(1,1,1,1,-1,-1,-1,-1,...)$ and so on, which is a basic sequence. Let $n$ be such that $f_1+...+f_n>\frac{2}{3} \|f\|_1$. Then, for any $m$ such that $2^m\ge n$, $\left<f,e_n\right>=f_1+...+f_n\pm f_{n+1}\pm f_{n+2}...>\frac{1}{3} \|f\|_1\not\to 0$.

If $E=C[0,1]$, taking variations of Schauder's basis shows that $F$ does not contain neither discrete measures, nor the Lebesgue measure. At the moment I am not even sure that $F$ is non-trivial.

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.