there exists a sequence $x_n$ such that $\| x_n \|=1$ and $x_n$ converges weakly to $0$.

Solution 1:

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 $x\in c_0$.