# Closed Convex Set in terms of Bounded Linear Functionals

## Theorem

Let $X$ be a normed vector space over $\R$.

Let $X^\ast$ be the normed dual of $X$.

Let $C$ be a closed convex subset of $X$.

Then:

$\ds C = \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$

## Proof

Clearly if $x \in C$ then:

$\ds \map f x \le \sup_{c \in C} \map f c$

for all $f \in X^\ast$, by the definition of supremum.

That is:

$\ds C \subseteq \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$

To show that:

$\ds \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c} \subseteq C$

we prove that if $x_0 \not \in C$ then:

$\ds x_0 \not \in \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$

The Rule of Transposition will then give us the result.

Let $x_0 \not \in C$.

From Finite Topological Space is Compact, we have that $\set {x_0}$ is compact.

Clearly $\set {x_0}$ is disjoint from $C$.

We can therefore apply Hahn-Banach Separation Theorem: Compact Convex Set and Closed Convex Set, which gives the existence of $f \in X^\ast$, $c \in \R$ and $\epsilon > 0$ such that:

$\map f {x_0} \le c - \epsilon < c + \epsilon \le \map f x$ for each $x \in C$.

From the definition of infimum, we therefore have:

$\ds c + \epsilon \le \inf_{x \in C} \map f x$

so that:

$\ds \map f {x_0} \le c - \epsilon < c + \epsilon \le \inf_{x \in C} \map f x$

Then, from Negative of Supremum is Infimum of Negatives, we have:

$\ds -\map f {x_0} \ge \epsilon - c > -\paren {c + \epsilon} \ge \sup_{x \in C} \paren {-\map f x}$

Setting $g = -f \in X^\ast$ we have:

$\ds \map g {x_0} > \sup_{x \in C} \map g x$

So:

$\ds x_0 \not \in \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$

as required.

Hence we obtain:

$\ds \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c} \subseteq C$

and:

$\ds C = \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$

$\blacksquare$