If $(x_n)$ converges weakly to $x$, then $x$ is in the closure of the span of the $x_n$ I need your help with this problem that I founded it in a lecture notes.
Then, the problem says:
Let $ X $ be a normed space. 
Show that if a sequence $ (x_n) _ {n \in \mathbb {N}} $ in $ X $ converges weakly to $ x $ then $ x\in Y $, where $ Y $ is the closure of the vector space generated by $\{x_n : n \in \mathbb{N} \} $.
 A: Suppose that $x\notin Y$.
Then, since $\{x\}$ is a compact convex set and $Y$ is a closed linear subspace (and therefore a closed convex set also), there exists a linear functional $h\in X^*$ such that $\langle h,x\rangle >0$ and $\langle h, x_i\rangle =0$ for every $i$.
It is a straightforward aplication of the Mazur-Eidelheit separation theorem, can also be proved using Hahn-Banach theorem.
Now observe that $\langle h,x_i\rangle =0<\langle h,x\rangle$ so we cannot have the convergence  $\langle h,x_i\rangle \rightarrow\langle h,x\rangle$, which is a contradiction since $x_i\rightarrow x$ in the weak sense. 
Q.E.D.
A: For a convex set in a normed space, the closure and the weak closure coincide. 
Let $S = \mathbb{sp}\{x_n\}$. Since $S$ is convex, we have $Y = \overline{S} = {\overline{S}}^w$ (the latter denoting the weak closure).
Since $x_n \to x$ weakly, we have $x \in {\overline{S}}^w$, hence $x \in \overline{S}$.
To see why $x$ must be in ${\overline{S}}^w$, suppose $x \notin {\overline{S}}^w$. Then there is a weak open set containing $x$ that does not intersect $S$. The open set must contain a weak neighborhood of the form $\{y \, |\,  |\phi_i(x-y)| < \epsilon, \,  \, i=1,...,n \}$. However, since $\phi_i(x_n) \to \phi(x)$, we quickly obtain a contradiction.
A: Hint: Proof by contradiction, using Hahn-Banach to produce a linear functional.
