# Gauge (Minkowski functional) of Convex Set

Theorem: Let $K$ be a convex set.

(i) $p_K(x)\leq 1$ if $x\in K$.

(ii) $p_K(x)<1$ if and only if $x$ is an interior point of $K$.

We need the following definition: $p_K(x)=\inf\{a\mid a>0,x/a\in K\}$. This $p_K$ is often known as a Gauge or Minkowski functional.

Proof:

(i) If $x\in K$, then $x/1\in K$, so $P_K(x)\leq 1$. This is the easy part. The converse holds here, if $x\notin K$, then $p_K(x)>1$.

(ii) We first prove the “only if” part. Assume $p_K(x)<1$. Suppose $x$ is not an interior point of $K$, i.e. there exists $y\in X$ such that for all $\epsilon>0$, $x+ty\notin K$ for some $|t|<\epsilon$.

We then have $p_K(x+ty)>1$. Combining with the subadditive property of the gauge, we have $1. Rearranging, we get $p_K(x)>1-p_K(ty)$. By considering the various possibilities of the sign of $t$, and using the positive homogeneity of the gauge, we can obtain a contradiction. For example, if $t>0$, $p_K(x)>1-tp_K(y)$. Since $t\to 0$ as $\epsilon\to 0$, this implies $p_K(x)\geq 1$, a contradiction.

Conversely, if $x$ is an interior point of $K$, for all $y$ there exists $\epsilon>0$ such that $x+ty\in K$ for all $|t|<\epsilon$.

We have $p_K(x+ty)\leq 1$ for all $y$, for all $|t|<\epsilon$. Since it is a “for all” quantifier, we can choose in particular $y=x$, $t>0$.

Then we have $p_K((1+t)x)\leq 1$, which leads to $(1+t)p_K(x)\leq 1$ and $p_K(x)\leq \frac{1}{1+t}<1$. ## Author: mathtuition88

http://mathtuition88.com

Posted on Categories math

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