Hahn-Banach Separation Theorem/Normed Vector Space/Real Case/Compact Convex Set and Closed 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 $A$ be a compact convex set.

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

Then there exists $f \in X^\ast$, $c \in \R$ and $\epsilon > 0$ such that:

$A \subseteq \set {x \in X : \map f x \le c - \epsilon}$

and:

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

That is:

there exists $f \in X^\ast$, $c \in \R$ and $\epsilon > 0$ such that $\map f a \le c - \epsilon < c + \epsilon \le \map f b$ for $a \in A$ and $b \in B$.

Proof

Let:

$\delta = \dfrac 1 4 \inf \set {\norm {a - b} : a \in A, \, b \in B}$

so that:

$\map d {A, B} = 4 \delta$

where $d$ is the metric induced by $\norm {\, \cdot \,}$ and $\map d {A, B}$ is the $d$-distance between $A$ and $B$.

From Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive, we have that $\delta > 0$.

Let:

$A_\delta = A + \map B { {\mathbf 0}_X, \delta}$

and:

$B_\delta = B + \map B { {\mathbf 0}_X, \delta}$

where $\map B { {\mathbf 0}_X, \delta}$ is the open ball of center ${\mathbf 0}_X$ and radius $\delta$.

From Sum of Set and Open Set in Topological Vector Space is Open, we have that $A_\delta$ and $B_\delta$ are open.

From Open Ball is Convex Set, we have that $\map B { {\mathbf 0}_X, \delta}$ is convex.

From Sum of Convex Sets in Vector Space is Convex it follows that $A_\delta$ and $B_\delta$ are convex.

It remains to show that $A_\delta$ and $B_\delta$ are disjoint.

Aiming for a contradiction, suppose that $x \in A_\delta \cap B_\delta$.

Then there exists $a \in A$, $b \in B$ and $u, v \in \map B { {\mathbf 0}_X, \delta}$ such that:

$x = a + u = b + v$

Then:

$a - b = v - u$

Then, we have:

\(\ds \norm {a - b}\)

\(=\)

\(\ds \norm {v - u}\)

\(\ds \)

\(\le\)

\(\ds \norm v + \norm u\)

\(\ds \)

\(=\)

\(\ds 2 \delta\)

This contradicts that:

$\map d {A, B} = 4 \delta$

So we must have $A_\delta \cap B_\delta = \O$.

Hence, Hahn-Banach Separation Theorem: Real Case: Open Convex Set and Convex Set can be applied, and there exists $f \in X^\ast$ and $c \in \R$ such that:

$A_\delta \subseteq \set {x \in X : \map f x < c}$

and:

$B_\delta \subseteq \set {x \in X : \map f x \ge c}$

For aesthetics, pick $v_0$ such that $\map f {v_0} = 1$.

Let $r \in \R$ be such that:

$0 < r < \dfrac \delta {\norm {v_0} }$

and $a \in A$.

Then we have:

$\norm {r v_0} < \delta$

so that:

$a + r v_0 \in A_\delta$

So, we have:

$\map f {a + r v_0} < c$

giving, from linearity:

$\map f a < c - r$

This inequality holds for all $a \in A$ and:

$0 < r < \dfrac \delta {\norm {v_0} }$

so we must have:

$\map f a \le c - \dfrac \delta {\norm {v_0} }$

for all $a \in A$.

Similarly, for all:

$0 < r < \dfrac \delta {\norm {v_0} }$

we have $b - r v_0 \in B_\delta$, and so:

$\map f {b - r v_0} \ge c$

and so:

$\map f b \ge c + r$

This gives:

$\map f b \ge c + \dfrac \delta {\norm {v_0} }$

for all $b \in B$.

So, setting:

$\epsilon = \dfrac \delta {\norm {v_0} }$

we have:

$A \subseteq \set {x \in X : \map f x \le c - \epsilon}$

and:

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

as required.

$\blacksquare$