Rate of convergence of weakly null sequences If $x_n$ is a normalized, weakly-null sequence in a Banach space, and $\epsilon_n\to 0$, does there exists a non-zero functional $f$ such that $|f(x_n)|<\epsilon_n$ for all $n$?  
 A: No, in fact in the Banach space $c_0$, for any sequence $\epsilon_n$ of positive numbers decreasing to $0$, we can choose a normalized weakly-null sequence $x_n$ such that for every nonzero bounded linear functional $f$, 
$\limsup_{n \to \infty} |f(x_n)|/\epsilon_n = \infty$.
Let $e_j$ be the $j$'th unit vector $(0,\ldots, 0, 1, 0, \ldots)$. Consider a sequence $x_n$ such that, for each positive integer $m$ such that $\epsilon_m < 1/2$, $x_{2^m+1}, \ldots, x_{2^{m+1}}$ consists of $\epsilon_m^{1/2}\left(\pm e_1 \pm e_2 \ldots \pm e_m\right) + e_{m+1}$
for all $2^m$ choices of signs.  It is easy to see that this is normalized and converges weakly to $0$.
But for any nonzero $f \in c_0^* = \ell_1$, take $m$ large enough that $\sum_{j=1}^{m+1} |f_j| > \|f\|/2$.  Then some $k$ with $2^m+1 \le k \le 2^{m+1}$ gives $f_j (x_k)_j$ all the same signs for $1 \le j \le m+1$, and thus for large enough $m$, $$\frac{|f(x_k)|}{\epsilon_k} \ge \frac{\epsilon_m^{1/2}}{\epsilon_k}\sum_{j=1}^{m+1} |f_j| > \frac{\epsilon_m^{1/2} \|f\|}{2 \epsilon_k}> \frac{\|f\|}{2 \epsilon_k^{1/2}} \to \infty$$
