Minkowski Functional of Open Convex Set is Bounded

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 there exists a real number $c > 0$ such that:


 * $0 \le \map {p_C} x \le c \norm x$

for each $x \in X$.

Proof
From the definition of the Minkowski functional, we have:


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

We have:


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

so:


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

from the definition of infimum.

If $x = 0$, we have:


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

for each $t > 0$, so:


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

so:


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

So the desired inequality holds for $x = 0$ for all real numbers $c > 0$.

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 $x \in C$.

In Minkowski Functional of Open Convex Set is Well-Defined, it is shown that:


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

So from the definition of infimum, we have:


 * $\ds \inf \set {t > 0 : t^{-1} x \in C} \le \frac {2 \norm x} \delta$

That is:


 * $\ds \map {p_C} x \le \frac 2 \delta \norm x$

Note that $\delta > 0$ depended only on $C$, so we have:


 * $\ds 0 \le \map {p_C} x \le \frac 2 \delta \norm x$

for all $x \in X$, hence the result.