Equivalence of reflexive and weakly compact In a normed space $X$ is there an equivalence between these two proposition?
$1)$ $X$ is reflexive;
$2)$ $B$, the unit ball of $X$, is weakly compact.
 A: Yes.
A proof of this theorem can be found in:

Marian Fabian, Petr Habala, Petr Hajek, Vicente Montesinos Santalucia, Jan Pelant, Vaclav Zizler.  Functional Analysis and Infinite-Dimensional Geometry.

See Theorem 3.31.
Google Books link
Edit: The referenced theorem assumes that $X$ is Banach; however, this automatically follows from either of conditions (1) and (2):


*

*Since $X^{**}$ is always complete, if $X$ is reflexive then it is complete (as noted in Matt E's comment).

*Suppose $B$ is weakly compact.  Let $\{x_n\}$ be Cauchy in $X$.  Cauchy sequences are bounded so by rescaling we may assume $\{x_n\} \subset B$.  By weak compactness, $\{x_n\}$ has a weak cluster point $x$.  Fix $\epsilon > 0$ and choose $N$ so large that $\|x_n - x_m\| < \epsilon$ for $n,m \ge N$.  Let $n \ge N$.  Now choose an arbitrary $f \in X^*$ with $\| f \| \le 1$.  As $x$ is a weak cluster point, there exists $m \ge N$ with $|f(x_m) - f(x)| < \epsilon$.  We also have $|f(x_m) - f(x_n)| \le \|x_m - x_n\| < \epsilon$.  Hence $|f(x_n) - f(x)| < 2 \epsilon$.  Taking the supremum over $f$ and using the Hahn-Banach theorem, we have $\|x_n - x\| < 2 \epsilon$.  Thus $x_n \to x$ in norm, and we have shown that $X$ is complete.
