Minkowski Functional of Balanced Convex Absorbing Set in Vector Space is Seminorm

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a vector space over $\GF$.

Let $A \subseteq X$ be a set that is balanced, convex and absorbing.

Let $\mu_A$ be the Minkowski functional of $A$.


Then $\mu_A$ is a seminorm.


Proof

Case 1: $\GF = \R$

From Balanced Set in Vector Space is Symmetric, $A$ is symmetric.

Hence by Minkowski Functional of Symmetric Convex Absorbing Set in Real Vector Space is Seminorm, $\mu_A$ is a seminorm in this case.

$\Box$

Case 2: $\GF = \C$

From Minkowski Functional of Convex Absorbing Set is Positive Homogeneous, $\mu_A$ is a sublinear functional.

Hence we have:

$\map {\mu_A} {x + y} \le \map {\mu_A} x + \map {\mu_A} y$ for all $x, y \in X$

and hence Seminorm Axiom $\text N 3$: Triangle Inequality.

We also have:

$\map {\mu_A} {r x} = r \map {\mu_A} x$ for all $r \in \hointr 0 \infty$ and $x \in X$.

Towards Seminorm Axiom $\text N 2$: Positive Homogeneity, we want to show that:

$\map {\mu_A} {r x} = \map {\mu_A} x$ for all $r \in \C$ with $\cmod r = 1$.

From the definition of a balanced set, we have:

$r A \subseteq A$

and since $\cmod {r^{-1} } = 1$, we have:

$r^{-1} A \subseteq A$

Then we have:

$r A \subseteq A \subseteq r A$

so that:

$A = r A$

Then we have:

\(\ds \map {\mu_A} {r x}\) \(=\) \(\ds \inf \set {t > 0 : r x \in t A}\) Definition of Minkowski Functional
\(\ds \) \(=\) \(\ds \inf \set {t > 0 : x \in t r^{-1} A}\)
\(\ds \) \(=\) \(\ds \inf \set {t > 0 : x \in t A}\) since $\cmod {r^{-1} } = 1$
\(\ds \) \(=\) \(\ds \map {\mu_A} x\) Definition of Minkowski Functional

Now take $r \in \C \setminus \hointl 0 \infty$ and $x \in X$.

We have:

\(\ds \map {\mu_A} {r x}\) \(=\) \(\ds \map {\mu_A} {\frac r {\cmod r} \cmod r x}\)
\(\ds \) \(=\) \(\ds \cmod r \map {\mu_A} {\frac r {\cmod r} x}\) Definition of Sublinear Functional
\(\ds \) \(=\) \(\ds \cmod r \map {\mu_A} x\) since $\cmod {\dfrac r {\cmod r} } = 1$

proving Seminorm Axiom $\text N 2$: Positive Homogeneity.

$\blacksquare$

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