Can we prove that in a convex space, Weakly closed=> weak* closed ??

Question

We know that weak* Topology is smaller than weak topology. So weak* closed sets are weakly closed. Banach Mazur theorem says "Strongly closed implies weakly closed if space is convex." can we expect this result in weak and weak * topology as well??

Answer 1

Recall that $c_0^{*}=l^{1}$ and $(l^{1})^{*}=l^{\infty}$. The set $C=\{(a_n)\in l^{1}: \sum_n a_n=1\}$ is convex, weakly closed but not $weak^{*}$ closed: the basis vectors $e_n$ are in this set, they converge to $0$ in $weak^{*}$ sense.

Answer 2

Let $X=l^\infty$, $C=c_0$. Then $X = (l_1)^*$. The point $x=(1,1,\dots)$ is in $l^\infty\setminus c_0$. However the sequence $$ x_n = (\underbrace{1,1,\dots,1}_{first\ n\ entries},0,0,\dots) $$ of elements in $c_0$ converges weak-star to $x$. Hence the closed subspace $c_0$ is not (sequentially) weak-star closed.