Minkowski Functional of Open Convex Set in Normed Vector Space is Well-Defined

Theorem

Let $\Bbb F \in \set {\R, \C}$.

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

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

Then, for each $x \in X$:

$\set {t > 0 : t^{-1} x \in C} \ne \O$

and so the Minkowski functional of $C$ is well-defined.

Proof

If $x = 0$, then:

$t^{-1} x \in C$

for all $t > 0$, so:

$\set {t > 0 : t^{-1} x \in C} = \openint 0 \infty$

Now take $x \ne 0$.

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

$\norm x < \delta$

we have $x \in C$.

Note that we have, from positive homogeneity:

$\ds \norm {\frac \delta {2 \norm x} x} = \frac \delta 2$

so:

$\dfrac \delta {2 \norm x} x \in C$

We therefore have:

$\dfrac {2 \norm x} \delta \in \set {t > 0 : t^{-1} x \in C}$

so in particular:

$\set {t > 0 : t^{-1} x \in C} \ne \O$

We also have that:

$t \ge 0$ for all $t \in \set {t > 0 : t^{-1} x \in C}$

So, from the Continuum Property:

$\set {t > 0 : t^{-1} x \in C}$ has an infimum.

So:

$\ds \inf \set {t > 0 : t^{-1} x \in C}$ is well-defined.

Since $x \in X$ was arbitrary, we have:

the Minkowski functional of $C$ is well-defined

$\blacksquare$