properties of Minkowski’s functional

Let $X$ be a normed space, $K$ convex subset of $X$ and $0$ belongs to the interior of $K$ .Then

1.

$\rho_{K}(x)\geq 0$ for all $x\in X$
2.

$\rho_{K}(0)=0$
3.

$\rho_{K}(\lambda x)=\lambda\rho_{K}(x)$ , for all $\lambda\geq 0$ and $x\in X$
4.

$\rho_{K}(x+y)\leq\rho_{K}(x)+\rho_{K}(y)$ for all $x,y\in K$
5.

$\{x\in X\colon\rho_{K}(x)<1\}\subset K\subset\{x\in X\colon\rho_{K}(x)\leq 1\}$
6.

$K^{0}=\{x\in X\colon\rho_{K}(x)<1\}$ where $K^{0}$ denotes the interior of $K$
7.

$\bar{K}=\{x\in X\colon\rho_{K}(x)\leq 1\}$ where $\bar{K}$ denotes the closure of $K$
8.

$Bd(K)=\{x\in X\colon\rho_{K}(x)=1\}$ where the $Bd(K)$ denotes the boundary of $K$ .

Minkowski’s functional is a useful tool to prove propositions and solve exercises. Let us see an example
Example Let $K$ be a convex subset of $X$ . Show that $Ex(K)\subset Bd(K)$ , where $Ex(K)$ denotes the set of extreme points of $K$ .
If $x\in Ex(K)$ then from this follows that $x\in 1K$ and $\rho_{K}(x)=1$ . Now we hypothesize that $\rho_{K}(x)<1$ then there is a real number $s$ such that $\rho_{K}(x)<s<1$ and so $\rho_{K}(\frac{x}{s})<1$ . Therefore we have that $x=s\frac{x}{s}+(1-s)0\in K$ , that contradicts to the fact that $x\in Ex(K).$

Title	properties of Minkowski’s functional
Canonical name	PropertiesOfMinkowskisFunctional
Date of creation	2013-03-22 15:45:04
Last modified on	2013-03-22 15:45:04
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	10
Author	georgiosl (7242)
Entry type	Theorem
Classification	msc 46B20