Minkowski Functional of Open Convex Set recovers Set

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

Let $C$ be an open convex subset of $X$ with $0 \in C$.

Let $p_C$ be the Minkowski functional for $C$.

Then:


 * $C = \set {x \in X : \map {p_C} x < 1}$

Proof
We first show that:


 * $\set {x \in X : \map {p_C} x < 1} \subseteq C$

Suppose that:


 * $x \in \set {x \in X : \map {p_C} x < 1}$

Then:


 * $\inf \set {t > 0 : t^{-1} x \in C} < 1$

from the definition of a Minkowski functional.

So, there exists $0 < \alpha < 1$ such that:


 * $\alpha^{-1} x \in C$

Since $C$ is convex and $0 \in C$, we have:


 * $\alpha \paren {\alpha^{-1} x} + \paren {1 - \alpha} \times 0 \in C$

so:


 * $x \in C$

So we have:


 * $\set {x \in X : \map {p_C} x < 1} \subseteq C$

Now we show that:


 * $C \subseteq \set {x \in X : \map {p_C} x < 1}$

Let:


 * $x \in C$

Since $C$ is open, there exists $\epsilon > 0$ such that for all $y \in X$ with:


 * $\norm {y - x} < \epsilon$

we have $y \in C$.

So, we have:


 * $\ds \paren {1 + \frac \epsilon 2} x \in C$

So:


 * $\ds \frac 2 {2 + \epsilon} \in \set {t > 0 : t^{-1} x \in C}$

So from the definition of infimum, we have:


 * $\ds \map {p_C} x \le \frac 2 {2 + \epsilon}$

So:


 * $\map {p_C} x < 1$

We therefore have:


 * $C \subseteq \set {x \in X : \map {p_C} x < 1}$

So, from the definition of set equality, we have:


 * $C = \set {x \in X : \map {p_C} x < 1}$