Hahn-Banach Separation Theorem/Hausdorff Locally Convex Space/Real Case/Compact Convex Set and Closed 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 compact convex set.

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

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

$\ds \sup_{x \mathop \in A} \map f x < \inf_{x \mathop \in B} \map f x$

Proof

Since $B$ is closed, $X \setminus B$ is open.

Let $x \in A$.

Since $A \cap B = \O$, we have $x \in X \setminus B$.

So there exists an open neighborhood $O_x$ of $x$ such that $O_x \subseteq X \setminus B$.

From Classification of Open Neighborhoods in Topological Vector Space, there exists an open neighborhood $U_x$ of ${\mathbf 0}_X$ such that $O_x = x + U_x$.

From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary $1$, there exists an symmetric open neighborhood $V_x$ of ${\mathbf 0}_X$ such that:

$V_x + V_x \subseteq U_x$

Now note that:

$\ds A \subseteq \bigcup_{x \mathop \in A} \paren {x + V_x}$

Since $A$ is compact, there exists $x_1, \ldots, x_n \in A$ such that:

$\ds A \subseteq \bigcup_{i \mathop = 1}^n \paren {x_i + V_{x_i} }$

Now let:

$\ds V' = \bigcap_{i \mathop = 1}^n V_{x_i}$

From the definition of a topology, we have that $V'$ is an open neighborhood of ${\mathbf 0}_X$.

From Standard Topology of Locally Convex Space has Local Basis of Balanced Convex Absorbing Sets, there exists a convex open neighborhood of ${\mathbf 0}_X$ $V$ contained in $V'$.

From Sum of Set and Open Set in Topological Vector Space is Open, we have that $A + V$ is open.

From Sum of Convex Sets in Vector Space is Convex, $A + V$ is additionally convex.

With view to apply Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Real Case: Open Convex Set and Convex Set, we will show that $\paren {A + V} \cap B = \O$.

Let $x \in A + V$.

Then $x = y + z$ for $y \in A$ and $z \in V$.

Since:

$\ds y \in \bigcup_{i \mathop = 1}^n \paren {x_i + V_{x_i} }$

there exists $1 \le j \le n$ and $y_j \in V_{x_j}$ such that:

$y = x_j + y_j$

Since $z \in V$, we have $z \in V_{x_j}$.

So, we have:

$y + z = x_j + y_j + z \in x_j + V_{x_j} + V_{x_j}$

Since $V_{x_j} + V_{x_j} \subseteq U_{x_j}$, we have that:

$x \in x_j + U_{x_j}$

By construction, we have that $\paren {x_j + U_{x_j} } \cap B = \O$, so we have $x \not \in B$.

So we obtain $\paren {A + V} \cap B = \O$.

From Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Real Case: Open Convex Set and Convex Set, there exists $f \in X^\ast$ and $c \in \R$ such that:

$\map f x < \inf_{y \mathop \in B} \map f y \le \map f b$ for all $x \in A + V$ and $b \in B$.

That is:

$\map f {a + v} < \inf_{y \mathop \in B} \map f y \le \map f b$ for all $a \in A$, $v \in V$ and $b \in B$.

Write:

$\ds c = \inf_{y \mathop \in B} \map f y$

for brevity.

Since for any $a_0 \in A$ we have:

$\map f {a_0} < \map f b$ for all $b \in B$

we certainly have that $f$ is not the constant ${\mathbf 0}_X$.

So there exists $v \in V$ 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:

$\dfrac v N \in V$

by the definition of a convergent sequence.

Now, for each $a \in A$ we have:

\(\ds \map f a\)

\(<\)

\(\ds \map f a + \map f v\)

\(\ds \)

\(=\)

\(\ds \map f {a + v}\)

Definition of Linear Functional

\(\ds \)

\(<\)

\(\ds c\)

\(\ds \)

\(=\)

\(\ds \inf_{y \in B} \map f y\)

From Continuous Function on Compact Space is Bounded, we have that $f$ is bounded on $A$ and attains its bounds.

That is, there exists some $a_\ast \in A$ such that:

$\map f {a_\ast} = \sup_{y \mathop \in A} \map f y$

Since $\map f {a_\ast} < c$, obtain the result:

$\ds \sup_{x \mathop \in A} \map f x < \inf_{x \mathop \in B} \map f x$

$\blacksquare$