Hahn-Banach Separation Theorem/Normed Vector Space/Complex Case/Open Convex Set and Convex Set

Theorem

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\C$.

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:

$A \subseteq \set {x \in X : \map \Re {\map f x} < c}$

and:

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

That is:

there exists $f \in X^\ast$ and $c \in \R$ such that $\map \Re {\map f a} < c \le \map \Re {\map f b}$ for each $a \in A$ and $b \in B$.

Let $\struct {X_\R, \norm {\, \cdot \,}_\R}$ be the realification of $X$ equipped with the restricted norm.

Applying Hahn-Banach Separation Theorem: Real Case: Open Convex Set and Convex Set to $\struct {X_\R, \norm {\, \cdot \,}_\R}$, there exists a bounded linear functional $g : X_\R \to \R$ and $c \in \R$ such that:

$A \subseteq \set {x \in X : \map g x < c}$

and:

$B \subseteq \set {x \in X : \map g x \ge c}$

$\map g x = \map \Re {\map f x}$

for each $x \in X$.

Then $f \in X^\ast$ satisfies:

$A \subseteq \set {x \in X : \map \Re {\map f x} < c}$

and:

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

as desired.

$\blacksquare$