Rudin's functional analysis theorem 3.21.

Question

Small proof of

Suppose $X$ is a topological vector space on which $X^*$ separates points. Suppose $A$ and $B$ are disjoint, nonempty, compact, convex sets in $X$. Then there exits $\Lambda \in X^*$ such that $$ \sup_{x \in A} \text{Re}\; \Lambda x < \inf_{y \in B} \text{Re}\; \Lambda y $$

Proof below:

Let $X_w$ be $X$ with its topology. The sets $A$ and $B$ area clearly compact in $X_w$. They're also closed in $X_w$ (because $X_w$ is a Hausdorff space).

Why exactly are $A$ and $B$ closed? How does it follow from the fact that $X_w$ is Hausdorff?

Since $X_w$ is locally convex...

Why is $X_w$ locally convex? Is it possible it follows from the fact that every weak neighborhood contains sets of the form

$$ V = \left\{ x : |\Lambda_i x | < r_i \;,\; 1 \leq i \leq n\right\} $$

Which would form a local base, and since such sets are convex then we have a locally convex space?

... (b) of theorem 3.4. can be applied to $X_w$ in place of $X$; it gives us a $\Lambda \in (X_w)^*$ that satisfies (1). But we saw in section 3.11 (as consequence of theorem 3.10) that $(X_w)^* = X^*$.

This might help: http://mathonline.wikidot.com/compact-sets-in-hausdorff-topological-spaces — user8469759, Feb 06 '19 at 16:38

score 1 · Answer 1 · answered Feb 06 '19 at 16:46

To answer your two questions:

It is a fact from general topology that compact subsets of Hausdorff spaces are closed (see for instance: this stack exchange question).
Theorem 3.10 in Functional Analysis states that if $X$ is a topological vector space and $X'$ is a separating family of linear functionals on $X$, then the topology induced by $X'$ on $X$ is locally convex. Since in the hypothesis to this theorem $X^*$ separates points, the weak topology it induces is thus locally convex.

I was just reading through theorem 3.10 again. – user8469759 Feb 06 '19 at 16:48 — user8469759, Feb 06 '19 at 16:48

Rudin's functional analysis theorem 3.21.

1 Answers1