# Minkowski functional

Let $(E,\|\cdot\|)$ be a normed vector space over $\mathbb R$ or $\mathbb C$. Then for $K\subseteq E$ the function $x\mapsto \inf\{\alpha > 0: x\in\alpha K\}$ is called a Minkowski functional and it is known that it is a seminorm if $K$ is convex, $x\in K$ and $|\lambda|=1$ implies $\lambda x\in K$ and for each $x\in E$ there is some $\alpha$ s.t. $x\in \alpha K$.

My question is what happens if we drop the convexity property, esp. if there is a difference between finite and infinite dimensional vector spaces.

• Drop convexity and you lose the triangle inequality. Oct 18 '15 at 11:53
• Could you elaborate this? E.g. with an example? As I used convexity (as well as condition 2) to show that the Minkowski functional is a seminorm in the above setting, that was although my guess, however I couldn't find an example and $K=\mathbb{S}^1 \cup \{0\}$ shows that it is not sufficient to be not convex to violate the triangle inequality. Oct 18 '15 at 12:08
• Consider the set of points in $\mathbb{R}^2$ whose coordinates satisfy $\lvert x\rvert^p + \lvert y\rvert^p \leqslant 1$ for some fixed $p \in (0,1)$. Then $\mu(e_1) = \mu(e_2) = 1$, and $\mu(e_1+e_2) = 2^{1/p} > 2$. If you only consider balanced sets, convexity is equivalent to having the triangle inequality. Oct 18 '15 at 12:29
• The whole time I just though about the case $\mathbb K = \mathbb C$... That was really dumb :P Oct 18 '15 at 12:32
• The same example works for $\mathbb{C}^2$ (and with the obvious adapations for arbitrary $\mathbb{K}^n$). You need a space of dimension $\geqslant 2$ for examples, though. Oct 18 '15 at 12:36

Let $E=\mathbb{R}^2$ and $K=\{0\}\cup\mathbb{S}^1\cup\{(2,0),(-2,0),(0,2),(0,-2)\}$. $K$ satisfies all conditions but convexity. Then $\parallel (1,0)\parallel_K=\parallel (0,1)\parallel_K=\tfrac{1}{2}$ and $\parallel (1,1)\parallel_K=\sqrt{2}\nleq\tfrac{1}{2}+\tfrac{1}{2}$, violating the triangle inequality.