0
$\begingroup$

Suppose $E$ is a convex subset of a locally convex space $X$. Then the weak closure $\overline{E}_w$ of $E$ is equal to its original closure $\overline{E}$.

The proof starts as follows

$\overline{E}_w$ is weakly closed, hence originally closed, so that $\overline{E} \subset \overline{E}_w$.

I don't get that inclusion, could anyone expound it please?

The proof also continues for the opposite inclusion, but I don't get that either

To obtain the poosite inclusion, choose $x_0 \in X, x_0 \notin \overline{E}$. Part (b) of the separation theorem 3.4 shows that there exists $\Lambda \in X^*$ and $\gamma \in \mathbb{R}$ such that, for every $x \in \overline{E}$ $$ Re \; \Lambda x_0 < \gamma < Re \; \Lambda x $$ The set $\left\{ x : Re \; \Lambda x < \gamma \right\}$ is therefore a weak neighborhood of $x_0$ that does not intersect $E$, thus $x_0 \notin \overline{E}_w$. This proves $\overline{E}_w \subset \overline{E}$

$\endgroup$
3
$\begingroup$

First, the weak topology is weaker than the original topology (it has fewer open and closed sets). This means that every weakly closed set is also a closed set.

Second, the closure of $E$ is the smallest closed set that contains $E$. Since the weak closure of $E$ contains $E$ and is closed, it must include the closure of $E$.

$\endgroup$
  • $\begingroup$ I know this might appear silly, but why the weak closure of $E$ contains $E$? For some reason this going back and forth between weak topology and original topology drives me crazy. $\endgroup$ – user8469759 Jan 10 at 12:06
  • $\begingroup$ the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$. $\endgroup$ – supinf Jan 10 at 12:07
  • 1
    $\begingroup$ How about the other inclusion? $\endgroup$ – user8469759 Jan 10 at 12:09
2
$\begingroup$

Second part: $\{x: \Re \Lambda x <\gamma\}$ is the inverse image under $\Lambda$ of $\{z\in \mathbb C: \Re z <\gamma\}$ which is open in $\mathbb C$. Since each continuous linear functional is continuous for the weak topology this inverse image is an open set in the weak topology. This open set contains $x_0$ but it contains no point of $E$ (because $\Re \Lambda x >\gamma$ for all $x \in E$). This implies that it cannot belong to the weak closure of $E$. Hence $x_o\notin \overline {E}$ implies $x_0$ does not belong to weak closure of $E$. In other words, weak closure of $E$ is a subset of $\overline {E}$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.