# 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. 1.

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

2. 2.

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

3. 3.

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

4. 4.

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

5. 5.

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

6. 6.

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

7. 7.

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

8. 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) 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 PropertiesOfMinkowskisFunctional 2013-03-22 15:45:04 2013-03-22 15:45:04 georgiosl (7242) georgiosl (7242) 10 georgiosl (7242) Theorem msc 46B20