# A weakly closed set in $X$ that remains weakly-star closed in $X^{**}$

I have $X$ a (non-reflexive) Banach space and $B\subset X$ a weakly closed convex subset.

I wonder under what additional conditions (other than weak compactness) $B$ remains weakly-star closed in $X^{**}$.

My take on this:- the canonical embedding $J:(X,w)\to (X^{**},w^*)$ is linear continuous and my question refers to finding on what sets $B$ is $J$ a closed map.

Another remark: - because $B$ is strongly closed in $X$ it is so in $X^{**}$. Combined with $B$ convex that yields that $B$ is weakly closed in $X^{**}$. It remains to get to the weak-star topology on $X^{**}$.

• $J(X)$ is always $w^\ast$-dense in $X^{\ast\ast}$: This follows from Hahn-Banach and $(X^{\ast\ast},w^\ast)^\ast=X^\ast$. Hence $J(X)$ is closed if and only if $X$ is reflxive. – Jochen Jun 20 '17 at 13:17
• Norm bounded weakly closed subsets are weakly compact, so it holds for those. Further I think it holds for sets of the form $B=(x^{*})^{-1}(V)$ where $V$ is a closed subspace of $\Bbb K$ that is not all of $\Bbb K$. Thus it also holds for arbitrary intersections of these. If you are interested in that I will write up a proof. – s.harp Jun 20 '17 at 14:10
• @s.harp I am not so sure about "Norm bounded weakly closed subsets are weakly compact". For example the unit ball is norm bounded and weakly closed but it is not weakly compact (the space is non-reflexive). Do you agree? – Neutral Element Jun 20 '17 at 14:16
• You are right, I forgot Banach Alaoglu was about the weak* topology. Also I think my other statement is not correct, even though I currently don't see why the proof doesn't work. I think the heat over here is making my brain rot. – s.harp Jun 20 '17 at 15:38