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
$\Box$
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 in Normed Vector Space recovers Set, we have:
- $\map {p_C} {v_0} \ge 1$
From Minkowski Functional of Open Convex Set in Normed Vector Space is Sublinear Functional, we have:
- $p_C$ is a Sublinear 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:
| \(\ds \map {f_0} {t v_0}\) | \(=\) | \(\ds t\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds t \map {p_C} {v_0}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map {p_C} {t v_0}\) | Definition of Sublinear Functional |
For $t < 0$, we have:
| \(\ds \map {f_0} {t v_0}\) | \(=\) | \(\ds t\) | ||||||||||||
| \(\ds \) | \(<\) | \(\ds 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map {p_C} {t v_0}\) | Definition of Minkowski Functional in Normed Vector Space |
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 in Normed Vector Space 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:
| \(\ds -\map f x\) | \(=\) | \(\ds \map f x\) | Definition of Linear Functional | |||||||||||
| \(\ds \) | \(\le\) | \(\ds c \norm {-x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds c \norm x\) | Norm Axiom $\text N 2$: Positive Homogeneity |
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:
| \(\ds \map f {v_0 + a - b}\) | \(=\) | \(\ds \map f {v_0} + \map f a - \map f b\) | Definition of Linear Functional | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map {f_0} {v_0} + \map f a - \map f b\) | since $f$ extends $f_0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 + \map f a - \map f b\) |
We have that:
- $v_0 + a - b \in C$
So, from Minkowski Functional of Open Convex Set in Normed Vector Space recovers Set, we have:
- $\map {p_C} {v_0 + a - b} < 1$
so:
| \(\ds \map f {v_0 + a - b}\) | \(\le\) | \(\ds \map {p_C} {v_0 + a - b}\) | ||||||||||||
| \(\ds \) | \(<\) | \(\ds 1\) |
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:
| \(\ds \map f {a' + \frac \epsilon {\norm {v_0} } v_0}\) | \(=\) | \(\ds \map f {a'} + \map f {\frac \epsilon {\norm {v_0} } v_0}\) | Definition of Linear Functional | |||||||||||
| \(\ds \) | \(=\) | \(\ds c + \map {f_0} {\frac \epsilon {\norm {v_0} } v_0}\) | since $f$ extends $f_0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds c + \frac \epsilon {\norm {v_0} }\) | ||||||||||||
| \(\ds \) | \(>\) | \(\ds c\) |
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 $b \in B$.
$\blacksquare$