# Theorem 3.18, Rudin's functional analysis

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.

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 $$\Lambda\in K$$.