Theorem: Hahn-Banach Separation Theorem/Normed Vector Space/Real Case/Compact Convex Set and Closed Convex Set

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.
{{AimForCont}} 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: 
{{begin-eqn}}
{{eqn	| l = \norm {a - b}
		| r = \norm {v - u}
}}
{{eqn	| o = \le
		| r = \norm v + \norm u
}}
{{eqn	| r = 2 \delta
}}
{{end-eqn}}
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.
{{qed}}
Category:Hahn-Banach Separation Theorem
Category:Convex Sets (Vector Spaces)
Category:Compact Normed Vector Spaces
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