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 $x^*: x^*(x_n)\rightarrow x^*(x)$).
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 $1$.