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 $n$? 



Rate of convergence of weakly null sequences
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 $m$, $$\frac{|f(x\_k)|}{\epsilon\_k} \ge \frac{\epsilon\_m^{1/2}}{\epsilon\_k}\sum\_{j=1}^{m+1} |f\_j| > \frac{\epsilon\_m^{1/2} \|f\|}{2 \epsilon\_k}> \frac{\|f\|}{2 \epsilon\_k^{1/2}} \to \infty$$


