Weak* closure of sets

functional-analysisweak-topology

I struggle with the following exercise about the weak* closure of a set. Weak and Weak* topology is a bit a weak (pun intended) spot of mine so i would like to ask if somebody could take a look, especially at Nr.2.

Let $X$ be a Banach space and $V\subseteq X$ and $W\subseteq X^*$ be linear subspaces, show:

  1. $V^\bot \subseteq X^*$ is weak* closed, and $^\bot W\subseteq X$ is closed.

This is clear. $^\bot W = \bigcap_{\lambda\in W}\ker(\lambda)$, $V^\bot = \bigcap_{\hat{x}\in\hat{V}}\ker(\hat{x})$, where $\hat{x}$ and $\hat{V}$ denote the canonical embedding into its double dual.
This part also holds for arbitrary sets.

  1. Given the first part show that $(^\bot W)^\bot$ is the weak* closure of $W$.

I assume this is the part where I have to use that $V$ and $W$ are linear subspaces. If I let $J$ denote the natural embedding of $X$ into in its bidual space $X^{\ast\ast}$ I can also find this unholy abomination of expression:
$$(^\bot W)^\bot=\bigcap_{\hat{x}\in J(\bigcap_{\lambda\in W}\ker(\lambda))}\ker(\hat{x})$$
This may already show the claim if could interpret it correctly.

  1. Show that $(^\bot W)^\bot = W$ if $W$ is weak* closed. Show that if $X$ is reflexive and $W$ is closed, then $(^\bot W)^\bot = W$.

The first one is clear. The weak* closure of a weak* closed set is itself. The second one, if $X$ is reflexive, then the weak and weak* topology are the same and thus the weak* closure and the weak closure are the same, and since $W$ is closed, its weakly closed implying $(^\bot W)^\bot = W$.

  1. Give an example of a non-reflexive space $X$ and a proper closed subspace $W\subseteq X^*$ such that $^\bot W = \{0\}$.

I haven't yet tried to find an example.

Thanks for having a look.

Best Answer

$V^\bot=\bigcap_{x\in V} \text{ev}_{x}^{-1}(0)$ where $\text{ev}_{x}$ denotes the evaluation map at $v$ which are continuous by definition of weak* topology.

Now the first part gives that $(^\bot W)^\bot$ is closed in the weak* toplogy and is a set which contains $W$. Hence the weak* closure of $W$ is contained in $(^\bot W)^\bot$ .

Now let $a\in \overline{W}$ where $\overline{W}$ denotes the weak* closure.

To show that $(^\bot W)^{\bot}\subseteq \overline{W}$ you need to do it by the Hahn Banach Theorem. Here is a link to a proof.

Hence, there exists some $x$ such that $f(g)=g(x)$ for all $g\in X^{*}$.

So $\overline{W}\subseteq (^\bot W)^{\bot}$

Hence $\overline{W}=(^\bot W)^{\bot}$

Now if $W$ itself is weak* closed, then by above you have $W=(^\bot W)^{\bot}$.

If $X$ was reflexive and $W$ is closed, then again you can show that $(^\bot W)^{\bot}=W^{cl}$ where $W^{cl}$ denotes the closure of $W$ in norm topology (the difference is that now you can work with sequences and derive things easily).

Related Question