Equivalence of reflexive and weakly compact

Question

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.

Answer 1

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):

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

2. 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.