a question regarding weak convergence How can I prove the following statement?
Let X be a normed vector space, $x_n$ a bounded sequence in X, x $\in$ X and X* the dual space of X.
To show: $x_n$ weakly converges to $x$ as  $n\rightarrow \infty$ if and only if there exists a subset $B \subseteq X^*$ with $X^*$ is the closure of $L(B)$ (the linear hull of $B$), such that $y^*(x_n) \rightarrow y^*(x)$ for every $y^* \in B$.  
(where $x_n$ converges weakly to $x$ iff for every $x^*: x^*(x_n)\rightarrow x^*(x)$).
 A: This is wrong. Let $X = l^2$ which is a separable Hilbert Space. Define $B:= \{e_n : n\in N\}$ and  take the sequence $\{ke_k\}_{k=1}^{\infty}$. Clearly for any $e_n \in B$ 
$$\left \langle e_n , k e_k \right \rangle \rightarrow  0  ,~ as ~ k \rightarrow \infty  $$
And $cl (L(B) )  = l^2 =X^*.$
But $\{ke_k\}_{k=1}^{\infty}$ is not weak convergent, since it is unbounded. 
P.S.   $e_n = (0,0,0...0,1,0,...0,0,0)$ the nth coordinate is $1$.
A: $\Rightarrow$ : Set $B=X^*$
$\Leftarrow$ : Let $B$ be a subset of $X^*$ with the properties that $X^* = l(L(B))$ and $ \forall y^*\in B :y^*(x_n)−>y^∗(x)$
T.p.: $z^∗(x_n) \rightarrow z^∗(x) \forall z^∗\in X^∗$. Let $z^∗ \in X^∗$be arbitrarily chosen. Because $cl(L(B))=X^∗: \exists a^∗$, which is a linear combination of elements in B, i.e.: $a^∗=\sum \limits_{i=1}^m \lambda_i * a_i^* ,a_i^* \in B \forall i.\\$ Let $\epsilon > 0$ be arbitrarily chosen.
Then, using the operator norm we get:
$\lVert z^∗(x_n)−z^∗(x) \rVert <=\lVert z^∗(x_n)−a^∗(x_n) \rVert +\lVert a^∗(x_n)−a^∗(x) \rVert+ \lVert a^∗(x)−z^∗(x) \rVert <= \\ \lVert z^∗−a^∗\rVert * \sup_n \lVert x_n \rVert +\lVert a^∗(x_n)−a^∗(x) \rVert +\lVert a^∗−z^∗\rVert ∗ \lVert x \rVert < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$.
The first and third term become arbitrarily small ($<\frac{\epsilon}{3}$) because $cl(L(B))=X^*$ and $x_n$ is bounded, the second one because it is given that holds $\forall a_i^* \in B$ and for $\epsilon > 0$ arbitrary:
 $\lVert a^∗(x_n)−a^∗(x) \rVert = \lVert a^∗(x_n - x) \rVert = \lVert \sum \limits_{i=1}^m \lambda_i * a_i^* (x_n-x) \rVert <= \sum \limits_{i=1}^m |\lambda_i| \lVert a_i^*(x_n - x) \rVert <= m*max_i|\lambda _i|*\lVert a_i^*(x_n - x) \rVert < \frac{\epsilon}{3}$, when $\lVert a_i^*(x_n - x) \rVert < \frac{\epsilon}{3*m*max_i|\lambda_i|} $.
Thanks a lot for helping me!
