Minkowski Functional of Open Convex Set in Normed Vector Space 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 in Normed Vector Space 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.
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $21.1$: The Minkowski Functional