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.



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> 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 $f\in F$?
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Criterion for weak convergence of sequences
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](http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10022.pdf)) 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| $ $\leq \delta (\|M\|+1/k)$.)


