Show that : $ x_n \overset {\sigma (X,X')}{\to} x $

Question

Let $ X $ be a separeble reflexive Banach space and $ (x_n)_n $ be a sequence of elements of $ X $ such that :

$(i)$. It admits a subsequence weakly converges to $ x \in X $

$(ii)$. Every limit point of $(x_n)_n $ must equal $x $.

Show that : $$ x_n \overset {\sigma (X,X')}{\to} x $$ An idea please

Answer 1

I doubt the question may be wrong.

Consider the case that $X=\mathbb{R}$ (with real scalar field). Clearly, the one dimensional Banach space $X$ is reflexive and separable.

Define $$ x_{n}=\begin{cases} n, & \mbox{if }n\mbox{ is odd}\\ 0, & \mbox{if }n\text{ is even} \end{cases}. $$ Clearly the subsequence $(x_{2n})$ converges to $x=0$ in norm, and trivially $x_{2n}\rightarrow x$ weakly. Moreover, $(x_{n})$ has only one limit point, namely $x=0$. That is, both condition s are satisfied. However, it is false that $x_{n}\rightarrow x$ weakly because if $x_{n}\rightarrow x$ weakly, then for any $f\in X^{\ast}=\mathbb{R}$, $fx_{n}=\langle f,x_{n}\rangle\rightarrow\langle f,x\rangle=fx$ (Here, $fx$ means the product of $f$ and $x$), which obviously does not hold.