Minkowski Functional of Open Convex Set is Well-Defined

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

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