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.


(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<p_K(x+ty)\leq p_K(x)+p_K(ty). 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

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