# Rudin's functional analysis theorem 3.12

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

## 2 Answers

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 $$E$$.

• 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. Jan 10 '19 at 12:06
• the weak closure is defined as the smallest weakly closed set that contains $E$. So by definition, it has to contain $E$. Jan 10 '19 at 12:07
• How about the other inclusion? Jan 10 '19 at 12:09

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 $$\overline {E}$$.

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