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Just a quick question about the the following theorem

In a locally convex space $X$, every weakly bounded set is originally bounded and viceversa.

Proof: Since every weak neighborhood of $0$ in $X$ is an original neighborhood of $0$, it is obvious from the definition of "bounded" that every originally bounded subset of $X$ is weakly bounded. The converse is the non trivial part of the theorem. Suppose $E \subset X$ is weakly bounded and $U$ is an original neighborhood of $0$ in $X$. Since $X$ is locally convex, there's a convex, balanced, original neighborhood $V$ of $0$ in $X$ such that $\overline{V} \subset U$. Let $K \subset X^*$ be the polar of $V$: $$ K = \left\{\Lambda \in X^* : |\Lambda x | \leq 1 \;\text{for all} \; x \in V \right\} \;\;\;\; (1) $$ We claim that $$ \overline{V} = \left\{x \in X : |\Lambda x| \leq 1 \; \text{for all} \; \Lambda \in K \right\} \;\;\; (2) $$ It is clear that $V$ is a subset of the right side of (2) and hence so is $\overline{V}$, since the right side of (2) is closed. Suppose $x_0 \in X$ but $x_0 \notin \overline{V}$. Theorem 3.7 (with $\overline{V}$ in place of $B$) then shows $\Lambda x_0 > 1$ for some $\Lambda \in K$. This proves (2)

It is clear to me the author is somehow using theorem 3.7 to reach a contradiction. The statement is below for convenience

Suppose $B$ is a convex, balanced, closed set in a locally convex space $X$, $x_0 \in X$, but $x_0 \notin B$. Then there exists $\Lambda \in X^*$ such that $|\Lambda x | \leq 1$ for all $x \in B$, but $\Lambda x_0 > 1$

How exactly is such theorem used to reach the conclusion that $\Lambda x_0 > 1$ for some $\Lambda \in K$? I'm sure it has something to do using the definitions of $K$ and $\overline{V}$ but I can't manage to put all the pieces together.

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Let's rewrite the conclusion of Theorem 3.7 with $B$ replaced by $\overline V$:

Suppose $x_0\in X$, but $x_0\notin\overline V$. Then there exists $\Lambda\in X^*$ such that $|\Lambda x|\leq 1$ for all $x\in\overline V$ but $\Lambda x_0>1$.

For the given $\Lambda$, if $x\in V$, then $x\in \overline V$, so that so by construction (i.e. by the conclusion of the theorem) we have $|\Lambda x|\leq 1$. Since $x\in V$ was arbitrary, we have $\Lambda\in K$.

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  • $\begingroup$ The constructions the author uses sometimes to prove the theorems always amaze me. $\endgroup$ – user8469759 Jan 10 at 16:21
  • $\begingroup$ Are you using Rudin's Functional Analysis? $\endgroup$ – Aweygan Jan 10 at 16:22
  • $\begingroup$ Yes, indeed I'm. $\endgroup$ – user8469759 Jan 10 at 16:27

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