Minkowski Functional of Open Convex Set in Normed Vector Space 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:

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

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $21.1$: The Minkowski Functional