Minkowski functional

In functional analysis, given a linear space X, a Minkowski functional is a device that uses the linear structure to introduce a topology on X.

Motivation

Example 1

Consider a normed vector space X, with the norm ||·||. Let K be the unit sphere in X. Define a function p : X → R by

$p(x) = \inf \left\{r > 0: x \in r K \right\}.$

One can see that $p(x) = \|x\|$, i.e. p is just the norm on X. The function p is a special case of a Minkowski functional.

Example 2

Let X be a vector space without topology with underlying scalar field K. Take φ ∈ X' , the algebraic dual of X, i.e. φ : X → K is a linear functional on X. Fix a > 0. Let the set K be given by

$K = \{ x \in X : | \phi(x) | \leq a \}.$

Again we define

$p(x) = \inf \left\{r > 0: x \in r K \right\}.$

Then

$p(x) = \frac{1}{a} | \phi(x) |.$

The function p(x) is another instance of a Minkowski functional. It has the following properties:

1. It is subadditive: p(x + y) ≤ p(x) + p(y),
2. It is homogeneous: for all α ∈ K, p(α x) = |α| p(x),
3. It is nonnegative.

Therefore p is a seminorm on X, with an induced topology. This is characteristic of Minkowski functionals defined via "nice" sets. There's a one-to-one correspondence between seminorms and the Minkowski functional given by such sets. What is meant precisely by "nice" is discussed in the section below.

Notice that, in contrast to a stronger requirement for a norm, p(x) = 0 need not imply x = 0. In the above example, one can take a nonzero x from the kernel of φ. Consequently, the resulting topology need not be Hausdorff.

Definition

The above examples suggest that, given a (complex or real) vector space X and a subset K, one can define a corresponding Minkowski functional

$p_K:X \rightarrow [0, \infty)$

by

$p_K (x) = \inf \left\{r > 0: x \in r K \right\},$

which is often called the gauge of K.

It is implicitly assumed in this definition that 0 ∈ K and the set {r > 0: x ∈ r K} is nonempty. In order for p_K to have the properties of a seminorm, additional restrictions must be imposed on K. These conditions are listed below.

1. The set K being convex implies the subadditivity of p_K.
2. Homogeneity, i.e. p_K(α x) = |α| p_K(x) for all α, is ensured if K is balanced, meaning α K ⊂ K for all |α| ≤ 1.

Convexity of K

A simple geometric argument that shows convexity of K implies subadditivity is as follows. Suppose for the moment that p_K(x) = p_K(y) = r. Then for all ε > 0, we have x, y ∈ (r + ε) K = K' . The assumption that K is convex means K' is also. Therefore ½ x + ½ y is in K' . By definition of the Minkowski functional p_K, one has

$p_K\left( \frac{1}{2} x + \frac{1}{2} y\right) \le r + \epsilon = \frac{1}{2} p_K(x) + \frac{1}{2} p_K(y) + \epsilon .$

But the left hand side is ½ p_K(x + y), i.e. the above becomes

$p_K(x + y) \le p_K(x) + p_K(y) + \epsilon, \quad \mbox{for all} \quad \epsilon > 0.$

This is the desired inequality. The general case p_K(x) > p_K(y) is obtained after the obvious modification.

Note Convexity of K, together with the initial assumption that the set {r > 0: x ∈ r K} is nonempty, implies that K is absorbent.

Balancedness of K

Notice that K being balanced implies that

$\lambda x \in r K \quad \mbox{if and only if} \quad x \in \frac{r}{|\lambda|} K.$

Therefore

$p_K (\lambda x) = \inf \left\{r > 0: \lambda x \in r K \right\} = \inf \left\{r > 0: x \in \frac{r}{|\lambda|} K \right\} = \inf \left\{ | \lambda | \frac{r}{ | \lambda | } > 0: x \in \frac{r}{|\lambda|} K \right\} = |\lambda| p_K(x).$

Related links

• Hadwiger's theorem
• Hugo Hadwiger
• Morphological image processing