Hahn-Banach Separation Theorem/Hausdorff Locally Convex Space/Real Case/Open Convex Set and Convex Set
Theorem
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\R$ equipped with its standard topology.
Let $X^\ast$ be the topological dual space of $\struct {X, \PP}$.
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
Fix $a_0 \in A$ and $b_0 \in B$.
From Sum of Set and Open Set in Topological Vector Space is Open, we have:
- $A - B$ is open.
From Translation of Open Set in Topological Vector Space is Open, we have:
- $A - B + \paren {b_0 - a_0}$ is open.
Note that since $A \cap B = \O$, we have $0 \notin A - B$.
Hence $b_0 - a_0 \notin A - B + \paren {b_0 - a_0}$.
From Sum of Convex Sets in Vector Space is Convex: Corollary, we have that $A - B$ is convex.
From Translation of Convex Set in Vector Space is Convex, we have that $A - B + \paren {b_0 - a_0}$ is convex.
Applying Open Convex Set in Hausdorff Locally Convex Space is Separated from Points outside Set by Continuous Linear Functional, there exists $f \in X^\ast$ such that:
- $\map f v < \map f {b_0 - a_0}$
for all $v \in A - B + \paren {b_0 - a_0}$.
That is:
- $\map f {a - b + b_0 - a_0} < \map f {b_0 - a_0}$
for each $a \in A$ and $b \in B$.
Since $f$ is linear, we have:
- $\map f a < \map f b$
for each $a \in A$ and $b \in B$.
Then we have:
- $\map f {a_0} < \map f b$
for all $b \in B$.
Hence:
- $\ds \inf_{b \mathop \in B} \map f b > -\infty$
Let:
- $\ds c = \inf_{b \mathop \in B} \map f b$
Then, we have:
- $\map f a \le c \le \map f b$
It remains to show that $\map f a < c$ for each $a \in A$.
Let $a \in A$.
Since $A$ is open, there exists an open neighborhood $V_x$ of ${\mathbf 0}_X$ such that $x + V_x$ is an open neighborhood of $x$ in $A$.
Now, since we have:
- $\map f {a_0} < \map f b$
for each $b \in B$, we in particular have that $f$ is not constant ${\mathbf 0}_X$.
Hence there exists $v \in X$ such that:
- $\map f v \ne 0$
Replacing $v$ by $-v$ if necessary, suppose that:
- $\map f v > 0$
From Multiple of Vector in Topological Vector Space Converges, we have:
- $\dfrac v n \to {\mathbf 0}_X$ as $n \to \infty$
So for some $N \in \N$, we have:
- $\dfrac v N \in V_x$
by the definition of a convergent sequence.
Set:
- $v' = \dfrac v N$
Then we have $\map f {v'} > 0$ and $x + v' \in A$.
We therefore have:
\(\ds \map f a\) | \(<\) | \(\ds \map f a + \map f {v'}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map f {a + v'}\) | Definition of Linear Functional | |||||||||||
\(\ds \) | \(\le\) | \(\ds c\) |
In particular, $\map f a < c$ for each $a \in A$.
Hence we have found $f \in X^\ast$ such that:
- $\map f a < c \le \map f b$ for each $a \in A$ and $b \in B$.
$\blacksquare$