I'm trying to find an example for the next assertion:
Let $E$ a Banach space, I want a set in the dual of E, $E'$, such that is convex and closed in the weak-* topology $\sigma(E',E)$, but the set have to be not closed in the weak topology for the dual $\sigma(E',E'')$. For this, I know that the space $E$ have to be not reflexive, but I cannot get an explicit example.
I'm very grateful with any answer, suggestion or book recommendation to construct the example.
Pdata: Sorry for my english.
Since the weak* topology on $X'$ is coarser than weak topology on $X'$ therefore every closed set in weak* topology in $X'$ is automatically closed in weak topology in $X'.$
The converse is not true if $X'$ is not reflexive. In fact in every non reflexive space $X'$ there exists closed bounded and convex set in weak topology in $X'$ such that is not closed in weak* topology in $X'.$ To see this observe that if $X'$ is not reflexive then there exist a linear functional $f: X' \to \mathbb{R} $ such that it is continuous in weak topology but it is not continuous in weak* topology in $X'.$ Then the set $\mbox{Ker}f =\{x\in X' : f(x) = 0 \}$ is closed in weak topology but it is not closed in weak* topology. Moreover the set $\mbox{Ker}f$ is dense in $X'$ with respect to the weak* topology. Now take $B=\{x:\in X' : ||x||\leq 1\}\cap\mbox{Ker} f$ the set is closed bounded and convex and hence weak closed in $X'.$ Suppose that $B$ is closed in weak* topology then it is equal to its closure in weak* topology which is equal to the closed unit ball of $X'$ but then we have that the unit ball is a subset of the set $\mbox{ker}f$ but this implies that $f$ is zero functional which is obviously impossible since $f$ was not continuous in weak* topology.