Criterion for weak convergence of sequences Let $E$ be a normed space and let $F\subset E^{*}$. It is known that $F$ is dense if and only if the restriction of $\sigma(E,F)$ on $B_E$ coincides with the weak topology.
Hence, if $F$ is dense and we have a bounded net $e_\alpha$ and $e\in E$ such that $\left<e_\alpha,f\right>\to \left<e,f\right>$, for every $f\in F$, then $e_\alpha\to e$ weakly. In particular, that works for sequences. However, density of $F$ seems to be too strong for sequences.

Is there a nice condition weaker than density of $F$ that would imply that every bounded sequence $e_n\in E$ converge weakly to $e\in E$ as long as $\left<e_n,f\right>\to \left<e,f\right>$, for every $f\in F$?

 A: In a nutshell, no, at least in the separable case.  Let $F\subseteq E^*$ be not norm dense, and with $F$ (norm-) separable.  By Hahn-Banach there is $M\in E^{**}$ which is non-zero and annihilates $F$.  Let $f_0\in E^*$ with $\langle M,f_0 \rangle=1$.
I shall use Helly's Lemma (which I have failed to find an online reference for; it follows from e.g. the principle of local reflexivity) which says that if $N\subseteq E^*$ is finite-dimensional and $M\in E^{**}$ then for $\epsilon>0$ we can find $x\in E$ with $\|x\|\leq \|M\|+\epsilon$ and $\langle M,f\rangle = \langle f,x\rangle$ for $f\in N$.
Let $(f_n)$ be a norm-dense sequence in $F$.  For each $n$ there is hence $x_n\in E$ with $\|x_n\| \leq \|M\|+1/n$, with $\langle M,f_0\rangle = \langle f_0,x_n\rangle$ and with $\langle M,f_k\rangle = \langle f_k,x_n\rangle$ for $k\leq n$.  Thus $(x_n)$ is not norm-null.  I shall show that $\langle f,x_n \rangle \rightarrow 0$ for each $f\in F$.
(This follows as $(x_n)$ is bounded and $(f_n)$ is dense in $F$.  To give the details, for $f\in F$ and $\delta>0$ there is $k$ with $\|f-f_k\|<\delta$ and so if $n\geq k$ then $|\langle f,x_n\rangle| $ $\leq |\langle f-f_k,x_n\rangle| + |\langle f_k,x_n\rangle| $ $= |\langle f-f_k,x_n\rangle| + |\langle M, f_k\rangle| $ $= |\langle f-f_k,x_n\rangle| $ $\leq \delta (\|M\|+1/k)$.)
