Let $X$ be a reflexive Banach space. Any closed convex bounded set is weakly compact. I know it is true. But, I can't find a reference. Anyone can help?
[Math] Any closed convex bounded set is weakly compact in a reflexive Banach space.
functional-analysis
Best Answer
This is the easy direction of Eberlein-Smulyan, and is a straight-forward application of the Banach-Alaogu theorem: If $X$ is reflexive, it suffices to show that the closed unit ball $B$ is weakly compact. To see this, consider the natural map $J: X\to X^{\ast\ast}$ to the double dual, and check that it induces a homeomorphism $$ J: (X,\sigma(X,X^{\ast})) \to (X^{\ast\ast}, \sigma(X^{\ast\ast},X^{\ast})) $$ where the topology on $X$ is the weak topology, and that of $X^{\ast\ast}$ is the weak-star topology (just take a basic open set on the RHS, use reflexivity and look at its preimage under $J$ to prove that $J$ is continuous. That $J$ is open is simply a matter of reversing that argument).
Now, by Banach-Alaoglu, the closed unit ball $B^{\ast\ast}$ in $X^{\ast\ast}$ is weak-star compact. Hence, $J^{-1}(B^{\ast\ast})$ is weakly compact in $X$. Now check that $B = J^{-1}(B^{\ast\ast})$.