there exists a sequence $x_n$ such that $\| x_n \|=1$ and $x_n$ converges weakly to $0$. Let $X$ be a reflexive Banach space of infinite dimension. 
a) Prove that there exists a sequence $x_n$ such that $\| x_n \|=1$ and $x_n$ converges weakly to $0$.
b) Let $x_n$ be a sequence such that $\forall f \in X' \quad \exists \lim\limits_{n\to\infty} f(x_n)<\infty$ .Prove that $x_n$ converges weakly.
c) Find a Banach space non riflexive where b is false.
My idea:
the weak closure of $S^1$ is the closed unit ball, thus $0\in$ the weak closure of $S^1$. But to prove that there exists a sequence in $S^1$ that converges weakly to 0 we must prove that the weak topology is metrizzable.Is it always true in riflexive spaces?
I would appreciate any idea on the solution.Thank you in advance.
 A: a) Let $\{z_n\}_{n\in\mathbb{N}}\subset S_X$ be such a sequence that $\Vert z_n-z_m\Vert\geq 1/2$ provided $n\neq m$. This sequence exist thanks to Riesz lemma. Since $X$ is reflexive its unit ball is weakly compact, so we can extract weakly convergent subsequence $\{z_{n(k)}\}_{k\in\mathbb{N}}$. Define $y_k=z_{n(2k+1)}-z_{n(2k)}$ for each $k\in\mathbb{N}$. Clearly $\{y_k\}_{k\in\mathbb{N}}$ weakly converges to $0$. Define $x_k=\Vert y_k\Vert^{-1} y_k$. By construction $1/2\leq\Vert y_k\Vert\leq 2$ for all $k\in\mathbb{N}$, then $\{x_k\}_{k\in\mathbb{N}}$ also weakly converges to $0$ and what is more $\Vert x_k\Vert=1$ for all $k\in\mathbb{N}$
b) Let $i:X\to X^{**}$ be the natural embedding into the second dual. From assumption the family of operators $\{i(x_n)\}_{n\in\mathbb{N}}\subset\mathcal{B}(X^*,\mathbb{C})$ is pointwise bounded family, so by Banach-Steinhaus theorem it is uniformly norm bounded by some constant $C>0$. By assumption we have well defined function $\varphi(f):=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}i(x_n)(f)$.
Clearly $\varphi$ is linear, but it is also bounded because $|\varphi(f)|
=\lim_{n\to\infty}|i(x_n)(f)|\leq\limsup_{n\to\infty}\Vert i(x_n)\Vert\Vert f\Vert\leq C\Vert f\Vert$. Thus $\varphi\in X^{**}$. Since $X$ is reflexive, we have $x\in X$ such that $\varphi=i(x)$. Now for all $f\in X^*$ we have $f(x)=i(x)(f)=\varphi(f)=\lim_{n\to\infty}f(x_n)$ i.e. $\{x_n\}_{n\in\mathbb{N}}$ weakly converges to $x$.
c) Let $X=c_0$ and $\{e_n:n\in\mathbb{N}\}$ be its natural basis. Define $x=\sum_{k=1}^n e_k$, then for all $f\in c_0^*\cong_1 \ell_1$ we have $\lim_{n\to\infty} f(x_n)=\sum_{n=1}^\infty f_n\in\mathbb{C}$, though there is no $x\in c_0$ such that $f(x)=\sum_{n=1}^\infty f_n$ for all $f\in c_0^*$. Indeed the last equality would imply $x(k)=1$ for all $k\in\mathbb{N}$ which is impossible since $x\in c_0$.
A: What you said is not true in general, but there's a way around it. Find a closed subspace $S \subset X$ which is separable (and it is reflexive also by a common result). To do this, pick $\{x_{n}\}_{n \in \mathbb{N}}$ linearly independent vectors in $X$ and consider the subspace spanned by them. This should be separable (why? considering rational scalings of all $\{x_{n}\}_{n \in \mathbb{N}}$). Now since $S$ is separable and reflexive, $S^{*}$ is separable (this is a theorem, not trivial), which shows that $B_{X} \cap S$ is metrizable. Apply your argument to $B_{X} \cap S$.
