The Minkowski Functional

The Minkowski Functional

Before we define the Minkowski functional we must first define what it means for a point in a subset of a vector space to be an internal point of that set.

 Definition: Let $X$ be a vector space and let $S \subseteq X$. A point $x_0 \in S$ is an Internal Point of $S$ if for every $x \in X$ there exists a $\delta > 0$ such that $x_0 + \lambda x \in U$ if $|\lambda| < \delta$.

We can now define the Minkowski functional.

 Definition: Let $X$ be a vector space and let $K \subset X$ be a convex subset of $X$ such that $0$ is an internal point of $K$. Then the Minkowski Functional on $K$ is the functional $P_K : X \to [0, \infty)$ defined for all $x \in X$ by $P_K(x) = \mathrm{inf} \{ t \geq 0 : x \in tK \}$.

For example, if $X$ is a normed vector space and $K$ is the closed unit ball of $X$, then:

(1)
\begin{align} \quad P_K(x) = \| x \| \end{align}
 Theorem 1: Let $X$ be a normed vector space and let $K \subset X$ be a convex subset of $X$ such that $0$ is an internal point of $K$. Then: a) $P_K$ is subadditive. b) $P_K(\lambda x) = |\lambda|P_K(x)$ for all $\lambda \in \mathbb{C}$ and for all $x \in X$.

We prove part (a) only. Part (b) is a left as an exercise.

• Proof of a): Clearly $P_K(0) = 0$, so let $x \in tK$ and $y \in sK$ be such that $x, y \neq 0$ and $s, t > 0$. Then:
(2)
\begin{align} \quad \frac{1}{s + t}(x + y) = \frac{t}{s + t} \frac{1}{t} x + \frac{s}{s + t} \frac{1}{s} y \end{align}
• Since $x \in tK$ and $y \in sK$ we have that $\frac{1}{t} x \in K$ and $\frac{1}{s} y \in K$. So by the convexity of $K$ we have that:
(3)
\begin{align} \quad \frac{1}{s + t}(x + y) \in K \end{align}
• Hence:
(4)
\begin{align} \quad P_K(x, y) \leq s + t \leq \mathrm{inf} \{ t \geq 0 : x \in tK \} + \mathrm{inf} \{ s \geq 0 : y \in sK \} = P_K(x) + P_K(y) \end{align}
• Therefore $P_K$ is subadditive. $\blacksquare$