Hahn-Banach Separation Theorem/Normed Vector Space/Real Case/Open Convex Set and Convex Set

Theorem
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$. Let $A \subseteq X$ be an open convex set.

Let $B \subseteq X$ be a convex set disjoint from $A$.

Then there exists $f \in X^\ast$ and $c \in \R$ such that:


 * $A \subseteq \set {x \in X : \map f x < c}$

and:


 * $B \subseteq \set {x \in X : \map f x \ge c}$

That is:


 * there exists $f \in X^\ast$ and $c \in \R$ such that $\map f a < c \le \map f b$ for each $a \in A$ and $b \in B$.

Proof
Let $a_0 \in A$ and $b_0 \in B$.

Let:


 * $v_0 = b_0 - a_0$

Let:


 * $C = v_0 + A - B = \set {v_0 + a - b : a \in A, \, b \in B}$

Lemma
Note that:


 * $v_0 \in C$




 * $v_0 + a - b = v_0$

for some $a \in A$ and $b \in B$.

That is:


 * $a = b$

So:


 * $v_0 \in C$




 * $A \cap B \ne \O$

Since $A$ and $B$ are disjoint, we therefore have:


 * $v_0 \not \in C$

Define:


 * $X_0 = \span \set {v_0}$

Then each $x_0 \in X_0$, there exists a unique $t \in \R$ such that:


 * $x_0 = t v_0$

From Linear Span is Linear Subspace, we have:


 * $X_0$ is a linear subspace of $X$.

Define $f_0 : X_0 \to \R$ by:


 * $\map {f_0} {t v_0} = t$

Let $p_C$ be the Minkowski functional of $C$.

From Minkowski Functional of Open Convex Set recovers Set, we have:


 * $\map {p_C} {v_0} \ge 1$

From Minkowski Functional of Open Convex Set is Minkowski Functional, we have:


 * $p_C$ is a Minkowski functional.

With view to apply the Hahn-Banach Theorem: Real Vector Space, we now show that:


 * $\map {f_0} {x_0} \le \map {p_C} {x_0}$

for each $x_0 \in X_0$.

For $t \ge 0$, we have:

For $t < 0$, we have:

So:


 * $\map {f_0} {x_0} \le \map {p_C} {x_0}$

for each $x_0 \in X_0$.

From Hahn-Banach Theorem: Real Vector Space, there exists a linear functional $f : X \to \R$ extending $f_0$ such that:


 * $\map f x \le \map {p_C} x$

for each $x \in X$.

We now check that $f \in X^\ast$.

That is, that $f$ is bounded.

From Minkowski Functional of Open Convex Set is Bounded, there exists $c > 0$ such that:


 * $\map {p_C} x \le c \norm x$

for each $x \in X$.

So we have:


 * $\map f x \le c \norm x$

We then have:

So we have:


 * $\size {\map f x} \le c \norm x$

for each $x \in X$.

So:


 * $f \in X^\ast$

We now show that there exists $c \in \R$ such that:


 * $\map f a \le c \le \map f b$

for each $a \in A$ and $b \in B$.

Let $a \in A$ and $b \in B$.

Then:

We have that:


 * $v_0 + a - b \in C$

So, from Minkowski Functional of Open Convex Set recovers Set, we have:


 * $\map {p_C} {v_0 + a - b} < 1$

so:

So we have:


 * $1 + \map f a - \map f b < 1$

So:


 * $\map f a < \map f b$

for each $a \in A$ and $b \in B$.

So for any fixed $a_0 \in A$, we have:


 * $\map f {a_0} < \map f b$

So:


 * $\set {b \in B : \map f b}$ is bounded below.

So, from the Continuum Property, we have that:


 * $\inf \set {b \in B : \map f b}$

exists.

If:


 * $c = \inf \set {b \in B : \map f b}$

Then:


 * $\map f a \le c \le \map f b$

for each $a \in A$ and $b \in B$.

It remains to show that in fact:


 * $\map f a < c$

suppose that there existed $a' \in A$ with:


 * $\map f {a'} = c$

Since $A$ is open, there exists $\epsilon > 0$ such that for all $a \in A$ with:


 * $\norm {a - a'} < \epsilon$

we have $a \in A$.

Note that since $v_0 \not \in C$, we have $v_0 \ne 0$.

In particular, $\norm {v_0} \ne 0$.

So, we have:


 * $\ds a' + \frac \epsilon {\norm {v_0} } v_0 \in A$

We then have:

which contradicts $\map f a \le c$ for each $a \in A$.

So, we have:


 * $\map f a < c \le \map f b$

for each $a \in A$ and $b \in B$.