A sequence converges weakly if and only if it converges in the weak-top. space The problem is:
Let $E$ be a normed space, let $x\in E$, let $\left(x_{n}\right)_{n\in\mathbb{N}}\subset E$
be a sequence. I need to show that $\left(x_{n}\right)$ converges
weakly to $x$ if and only if $\left(x_{n}\right)$ converges to $x$
in $\left(E,\,\sigma\left(E,\, E^{*}\right)\right)$, $\sigma\left(E,\, E^{*}\right)$ denoting the weak topology on $E$.
My efforts:
I have that $x_{n}\rightarrow x$ in $\left(E,\,\sigma\left(E,\, E^{*}\right)\right)$
is equivalent to:
$\forall U\in\sigma\left(E,\, E^{*}\right),\, x\in U\,:\quad\left\{ n\in\mathbb{N}:\, x_{n}\notin U\right\}$ is finite.
My question:
It is unclear to me how to pass from the weak convergence in $E$ to the convergence in the weak topology on $E$.
How can I prove both directions?
Thanks, Franck.
 A: WLOG, set $x=0$ (weak translation is continuous):
We have to show that $x_{n}\rightharpoonup0\Leftrightarrow x_{n}\rightarrow0$
in $\left(E,\sigma\left(E,E^{\star}\right)\right).$
$$
\Rightarrow\forall\epsilon\quad\exists N_{\epsilon}\in\mathbb{N}\quad\forall f\in E^{\star}:\quad\left|f(x_{n})\right|<\epsilon\quad\forall n\geq N_{\epsilon}.
$$
Let $U\left(f_{1},...,f_{k};\epsilon\right)$ denote a weak neighborhood
of the origin:
$$
U\left(f_{1},...,f_{k};\epsilon\right):=\left\{ x\in E:\,\left|f_{i}\left(x\right)\right|<\epsilon,\, i=1,...,k\right\} .
$$
So $x_{n}\in U\left(f_{1},...,f_{k};\epsilon\right)$, except at most
for a finite number of indices. This is enough to infer that $x_{n}\rightarrow0$
in $\left(E,\sigma\left(E,E^{\star}\right)\right)$, since every weakly
open set $U$ containing the origin of $E$ contains a neighborhood
of the form $U\left(f_{1},...,f_{k};\epsilon\right)$ for some $f_{1},...,f_{k}\in E^{\star}$
and $\epsilon>0$.
For the reverse:
$x_{n}\rightarrow0$ in $\left(E,\sigma\left(E,E^{\star}\right)\right)\Rightarrow x_{n}\in U\left(f_{1},...,f_{k};\epsilon\right)$,
except for at most a finite number of indices, $f_{i}\in E^{\star},\,\epsilon>0$.
$\Rightarrow f_{i}(x_{n})\rightarrow0$ for $f_{i}\in E^{\star},\, i=1,...,k$.
Set $k=1\Rightarrow\forall f\in E^{\star}:\, f\left(x_{n}\right)\rightarrow f\left(0\right)=0\Rightarrow x_{n}\rightharpoonup0$
$\square$
