Prove that every reflexive Banach space is weakly complete. Here is the question:
A sequence $(x_{n})$ in a normed linear space $X$ is weakly Cauchy if $(Tx_{n})$ is a Cauchy sequence for every $T \in X^*.$ The space $X$ is weakly complete if every weakly Cauchy sequence in $X$ is weakly convergent. Prove that every reflexive Banach space is weakly complete. Could anyone help me in this proof please?
 A: Suppose $(x_n)$ is a weakly Cauchy sequence and let $f\in X^*$. Then $f(x_n)$ is a Cauchy sequence and by completeness of $\mathbb{C}$ it converges to some element $\alpha(f)\in\mathbb{C}$. It is easy to see that $\alpha$ is a linear functional $X^*\to\mathbb{C}$. It is a bit more tricky so show this functional is bounded. For that we identify the elements of $X$ with elements in $X^{**}$. So for each $f\in X^*$ we have $x_n(f)=f(x_n)\to\alpha(f)$. Hence for each $f\in X^*$ the sequence $x_n(f)$ is bounded by some constant $C(f)$. But now by the uniform boundedness principle the sequence $(x_n)$ itself must be bounded in $X^{**}$. Since $||x_n||_{X^{**}}=||x_n||$ we conclude that the sequence $(x_n)$ is bounded in $X$ by some $M>0$. Hence for each $f\in X^*$:
$|f(x_n)|\leq M\times||f||$
By passing to the limit we get $|\alpha(f)|\leq M||f||$. So $\alpha$ is indeed bounded, hence in $X^{**}$. But since $X$ is reflexive we conclude that $\alpha\in X$. And by definition for each $f\in X^*$ we have:
$f(x_n)\to \alpha(f)=f(\alpha)$
So indeed $x_n$ converges weakly to $\alpha$. 
A: Let $\{x_n\}$ is weakly Cauchy, i.e. $\forall\varphi\in X^\ast$ $|\varphi(x_n)-\varphi(x_m)|\to0$ by $n,m\to\infty$. So the numerical sequence $\{\varphi(x_n)\}$ is Caushy and it's converges to some number, say $a$. Since the convergent sequence is always bounded, then $\forall\varphi\in X^\ast$ a sequence $\{\varphi(x_n)\}$ is bounded. This means that $\{x_n\}$ is weakly-bounded in $X$. It's easy to see that boundedness $\{x_n\}$ follows from here (if $\{x_n\}$ is unbounded then we can assume that $\|x_n\|>n^2$. But $\forall\varphi\in X^\ast$, due the weak-boundedness $\varphi(x_n/n)\to0$ -- a contradiction). By the reflexivity, this means that for a sequence $\{x_n\}$ one can find a weakly convergent subsequence $\{x_{n_k}\}$, i.e. $x_{n_k}\rightharpoonup x_0\in X$ or $\forall\varphi\in X^\ast$ $\varphi(x_{n_k})\to\varphi(x_0)$. Due of the uniqueness of the limit, we obtain $a=\varphi(x_0)$, i.e. $x_{n}\rightharpoonup x_0$.
