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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 $\overline{E}_w \subset \overline{E}$

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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 $E$.

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  • $\begingroup$ 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. $\endgroup$ Jan 10 '19 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 '19 at 12:07
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    $\begingroup$ How about the other inclusion? $\endgroup$ Jan 10 '19 at 12:09
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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 $\overline {E}$.

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  • $\begingroup$ Why is $\{ x: \mathcal{R} \Lambda x<y\}$ the inverse image under $\Lambda$ of $\{ z \in \mathbb{R} : \mathcal{R}z<y\}$? Thanks! $\endgroup$ Dec 6 '20 at 2:18
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    $\begingroup$ That is just the definition of inverse image: $\Lambda^{-1}(S)$ is defined as $\{x: \Lambda x\in S\}$. @cciirrcclllee $\endgroup$ Dec 6 '20 at 4:37

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