Minkowski functional

Definition

Let K be a subset of a real or complex vector space X. Define the gauge of K or the Minkowski functional associated with or induced by K as being the function p_K : X \to [0, \infty], valued in the extended real numbers, defined by

p_K(x) := \inf {r 0 : x \in r K},

(recall that the infimum of the empty set is ,\infty, that is, \inf \varnothing = \infty). Here, {r 0 : x \in r K} is shorthand for {r \in \R : r 0 \text{ and } x \in r K}.

For any x \in X, p_K(x) \neq \infty if and only if {r 0 : x \in r K} is not empty. The arithmetic operations on \R can be extended to operate on \pm \infty, where \frac{r}{\pm \infty} := 0 for all non-zero real - \infty The products 0 \cdot \infty and 0 \cdot - \infty remain undefined.

Some conditions making a gauge real-valued

In the field of convex analysis, the map p_K taking on the value of ,\infty, is not necessarily an issue. However, in functional analysis p_K is almost always real-valued (that is, to never take on the value of ,\infty,), which happens if and only if the set {r 0 : x \in r K} is non-empty for every x \in X.

In order for p_K to be real-valued, it suffices for the origin of X to belong to the algebraic interior or core of K in X. If K is absorbing in X, where recall that this implies that 0 \in K, then the origin belongs to the algebraic interior of K in X and thus p_K is real-valued. Characterizations of when p_K is real-valued are given below.

Motivating examples

Example 1

Consider a normed vector space (X, |,\cdot,|), with the norm |,\cdot,| and let U := {x\in X : |x| \leq 1} be the unit ball in X. Then for every x \in X, |x| = p_U(x). Thus the Minkowski functional p_U is just the norm on X.

Example 2

Let X be a vector space without topology with underlying scalar field \mathbb{K}. Let f : X \to \mathbb{K} be any linear functional on X (not necessarily continuous). Fix a 0. Let K be the set K := {x \in X : |f(x)| \leq a} and let p_K be the Minkowski functional of K. Then p_K(x) = \frac{1}{a} |f(x)| \quad \text{ for all } x \in X. The function p_K has the following properties:

1. It is subadditive: p_K(x + y) \leq p_K(x) + p_K(y).
2. It is absolutely homogeneous: p_K(s x) = |s| p_K(x) for all scalars s.
3. It is nonnegative: p_K \geq 0.

Therefore, p_K is a seminorm on X, with an induced topology. This is characteristic of Minkowski functionals defined via "nice" sets. There is 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_K(x) = 0 need not imply x = 0. In the above example, one can take a nonzero x from the kernel of f. Consequently, the resulting topology need not be Hausdorff.

Common conditions guaranteeing gauges are seminorms

To guarantee that p_K(0) = 0, it will henceforth be assumed that 0 \in K.

In order for p_K to be a seminorm, it suffices for K to be a disk (that is, convex and balanced) and absorbing in X, which are the most common assumption placed on K.

If K is an absorbing disk in a vector space X then the Minkowski functional of K, which is the map p_K : X \to 0, \infty) defined by p_K(x) := \inf {r 0 : x \in r K}, is a seminorm on X. Moreover, p_K(x) = \frac{1}{\sup {r 0 : r x \in K}}.

More generally, if K is convex and the origin belongs to the [algebraic interior of K, then p_K is a nonnegative sublinear functional on X, which implies in particular that it is subadditive and positive homogeneous. If K is absorbing in X then p_{[0, 1] K} is positive homogeneous, meaning that p_{[0, 1] K}(s x) = s p_{[0, 1] K}(x) for all real s \geq 0, where [0, 1] K = {t k : t \in [0, 1], k \in K}. If q is a nonnegative real-valued function on X that is positive homogeneous, then the sets U := {x \in X : q(x) and D := {x \in X : q(x) \leq 1} satisfy [0, 1] U = U and [0, 1] D = D; if in addition q is absolutely homogeneous then both U and D are balanced.

Gauges of absorbing disks

Arguably the most common requirements placed on a set K to guarantee that p_K is a seminorm are that K be an absorbing disk in X. Due to how common these assumptions are, the properties of a Minkowski functional p_K when K is an absorbing disk will now be investigated. Since all of the results mentioned above made few (if any) assumptions on K, they can be applied in this special case.

Assume that K is an absorbing subset of X. It is shown that:

1. If K is convex then p_K is subadditive.
2. If K is balanced then p_K is absolutely homogeneous; that is, p_K(s x) = |s| p_K(x) for all scalars s.

Convexity and subadditivity

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 e 0, x, y \in K_e := (r, e) K. Since K is convex and r + e \neq 0, K_e is also convex. Therefore, \frac{1}{2} x + \frac{1}{2} y \in K_e. By definition of the Minkowski functional p_K, p_K\left(\frac{1}{2} x + \frac{1}{2} y\right) \leq r + e = \frac{1}{2} p_K(x) + \frac{1}{2} p_K(y) + e.

But the left hand side is \frac{1}{2} p_K(x + y), so that p_K(x + y) \leq p_K(x) + p_K(y) + 2 e.

Since e 0 was arbitrary, it follows that p_K(x + y) \leq p_K(x) + p_K(y), which is the desired inequality. The general case p_K(x) p_K(y) is obtained after the obvious modification.

Convexity of K, together with the initial assumption that the set {r 0 : x \in r K} is nonempty, implies that K is absorbing.

Balancedness and absolute homogeneity

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).

Algebraic properties

Let X be a real or complex vector space and let K be an absorbing disk in X.

• p_K is a seminorm on X.
• p_K is a norm on X if and only if K does not contain a non-trivial vector subspace.
• p_{s K} = \frac{1}{|s|} p_K for any scalar s \neq 0.
• If J is an absorbing disk in X and J \subseteq K then p_K \leq p_J.
• If K is a set satisfying {x \in X : p(x) then K is absorbing in X and p = p_K, where p_K is the Minkowski functional associated with K; that is, it is the gauge of K.
• In particular, if K is as above and q is any seminorm on X, then q = p if and only if {x \in X : q(x)
• If x \in X satisfies p_K(x) then x \in K.

Topological properties

Assume that X is a (real or complex) topological vector space (not necessarily Hausdorff or locally convex) and let K be an absorbing disk in X. Then

\operatorname{Int}_X K ; \subseteq ; {x \in X : p_K(x)

where \operatorname{Int}_X K is the topological interior and \operatorname{Cl}_X K is the topological closure of K in X. Importantly, it was not assumed that p_K was continuous nor was it assumed that K had any topological properties.

Moreover, the Minkowski functional p_K is continuous if and only if K is a neighborhood of the origin in X. If p_K is continuous then \operatorname{Int}_X K = {x \in X : p_K(x)

Minimal requirements on the set

This section will investigate the most general case of the gauge of any subset K of X. The more common special case where K is assumed to be an absorbing disk in X was discussed above.

Properties

All results in this section may be applied to the case where K is an absorbing disk.

Throughout, K is any subset of X.

Suppose that K is a subset of a real or complex vector space X.

1. Strict positive homogeneity: p_K(r x) = r p_K(x) for all x \in X and all positive real r 0.

• Positive/Nonnegative homogeneity: p_K is nonnegative homogeneous if and only if p_K is real-valued.
  • A map p is called nonnegative homogeneous if p(r x) = r p(x) for all x \in X and all nonnegative real r \geq 0. Since 0 \cdot \infty is undefined, a map that takes infinity as a value is not nonnegative homogeneous.

1. Real-values: (0, \infty) K is the set of all points on which p_K is real valued. So p_K is real-valued if and only if (0, \infty) K = X, in which case 0 \in K.

• Value at 0: p_K(0) \neq \infty if and only if 0 \in K if and only if p_K(0) = 0.
• Null space: If x \in X then p_K(x) = 0 if and only if (0, \infty) x \subseteq (0, 1) K if and only if there exists a divergent sequence of positive real numbers t_1, t_2, t_3, \cdots \to \infty such that t_n x \in K for all n. Moreover, the zero set of p_K is \ker p_K \stackrel{\scriptscriptstyle\text{def}}{=} \left{y \in X : p_K(y) = 0 \right} = {\textstyle\bigcap\limits_{e 0}} (0, e) K.

1. Comparison to a constant: If 0 \leq r \leq \infty then for any x \in X, p_K(x) if and only if x \in (0, r) K; this can be restated as: If 0 \leq r \leq \infty then p_K^{-1}([0, r)) = (0, r) K.

• It follows that if 0 \leq R is real then p_K^{-1}([0, R]) = {\textstyle\bigcap\limits_{e 0}} (0, R + e) K, where the set on the right hand side denotes {\textstyle\bigcap\limits_{e 0}} [(0, R + e) K] and not its subset \left[{\textstyle\bigcap\limits_{e 0}} (0, R + e)\right] K = (0, R] K. If R 0 then these sets are equal if and only if K contains \left{y \in X : p_K(y) = 1 \right}.
• In particular, if x \in R K or x \in (0, R] K then p_K(x) \leq R, but importantly, the converse is not necessarily true. R = 0 then (0, R] K = \varnothing so x \in (0, R] K implies R \neq 0. ---

1. Gauge comparison: For any subset L \subseteq X, p_K \leq p_L if and only if (0, 1) L \subseteq (0, 1) K; thus p_L = p_K if and only if (0, 1) L = (0, 1) K.

• The assignment L \mapsto p_L is order-reversing in the sense that if K \subseteq L then p_L \leq p_K.
• Because the set L := (0, 1) K satisfies (0, 1) L = (0, 1) K, it follows that replacing K with p_K^{-1}([0, 1)) = (0, 1) K will not change the resulting Minkowski functional. The same is true of L := (0, 1] K and of L := p_K^{-1}([0, 1]).
• If D \stackrel{\scriptscriptstyle\text{def}}{=} \left{y \in X : p_K(y) = 1 \text{ or } p_K(y) = 0 \right} then p_D = p_K and D has the particularly nice property that if r 0 is real then x \in r D if and only if p_D(x) = r or p_D(x) = 0.It is in general false that x \in r D if and only if p_D(x) = r (for example, consider when p_K is a norm or a seminorm). The correct statement is: If 0 then x \in r D if and only if p_D(x) = r or p_D(x) = 0. Moreover, if r 0 is real then p_D(x) \leq r if and only if x \in (0, r] D.

1. Subadditive/Triangle inequality: p_K is subadditive if and only if (0, 1) K is convex. If K is convex then so are both (0, 1) K and (0, 1] K and moreover, p_K is subadditive.
2. Scaling the set: If s \neq 0 is a scalar then p_{s K}(y) = p_K\left(\tfrac{1}{s} y\right) for all y \in X. Thus if 0 is real then p_{r K}(y) = p_K\left(\tfrac{1}{r} y\right) = \tfrac{1}{r} p_K(y).
3. Symmetric: p_K is symmetric (meaning that p_K(- y) = p_K(y) for all y \in X) if and only if (0, 1) K is a symmetric set (meaning that(0, 1) K = - (0, 1) K), which happens if and only if p_K = p_{- K}.
4. Absolute homogeneity: p_K(u x) = p_K(x) for all x \in X and all unit length scalars uu is having unit length means that |u| = 1. if and only if (0, 1) u K \subseteq (0, 1) K for all unit length scalars u, in which case p_K(s x) = |s| p_K(x) for all x \in X and all non-zero scalars s \neq 0. If in addition p_K is also real-valued then this holds for all scalars s (that is, p_K is absolutely homogeneousThe map p_K is called absolutely homogeneous if |s| p_K(x) is well-defined and p_K(s x) = |s| p_K(x) for all x \in X and all scalars s (not just non-zero scalars).).

• (0, 1) u K \subseteq (0, 1) K for all unit length u if and only if (0, 1) u K = (0, 1) K for all unit length u.
• s K \subseteq K for all unit scalars s if and only if s K = K for all unit scalars s; if this is the case then (0, 1) K = (0, 1) s K for all unit scalars s.
• The Minkowski functional of any balanced set is a balanced function.

1. Absorbing: If K is convex or balanced and if (0, \infty) K = X then K is absorbing in X.

• If a set A is absorbing in X and A \subseteq K then K is absorbing in X.
• If K is convex and 0 \in K then [0, 1] K = K, in which case (0, 1) K \subseteq K.

1. Restriction to a vector subspace: If S is a vector subspace of X and if p_{K \cap S} : S \to [0, \infty] denotes the Minkowski functional of K \cap S on S, then p_K\big\vert_S = p_{K \cap S}, where p_K\big\vert_S denotes the restriction of p_K to S.

The proofs of these basic properties are straightforward exercises so only the proofs of the most important statements are given.

The proof that a convex subset A \subseteq X that satisfies (0, \infty) A = X is necessarily absorbing in X is straightforward and can be found in the article on absorbing sets.

For any real t 0,

{r 0 : t x \in r K} = {t(r/t) : x \in (r/t) K} = t {s 0 : x \in s K}

so that taking the infimum of both sides shows that

p_K(tx) = \inf {r 0 : t x \in r K} = t \inf {s 0 : x \in s K} = t p_K(x).

This proves that Minkowski functionals are strictly positive homogeneous. For 0 \cdot p_K(x) to be well-defined, it is necessary and sufficient that p_K(x) \neq \infty; thus p_K(tx) = t p_K(x) for all x \in X and all non-negative real t \geq 0 if and only if p_K is real-valued.

The hypothesis of statement (7) allows us to conclude that p_K(s x) = p_K(x) for all x \in X and all scalars s satisfying |s| = 1. Every scalar s is of the form r e^{i t} for some real t where r := |s| \geq 0 and e^{i t} is real if and only if s is real. The results in the statement about absolute homogeneity follow immediately from the aforementioned conclusion, from the strict positive homogeneity of p_K, and from the positive homogeneity of p_K when p_K is real-valued. \blacksquare

Examples

1. If \mathcal{L} is a non-empty collection of subsets of X then p_{\cup \mathcal{L}}(x) = \inf \left{p_L(x) : L \in \mathcal{L} \right} for all x \in X, where \cup \mathcal{L} \stackrel{\scriptscriptstyle\text{def}}{=} {\textstyle\bigcup\limits_{L \in \mathcal{L}}} L.

• Thus p_{K \cup L}(x) = \min \left{p_K(x), p_L(x) \right} for all x \in X.

1. If \mathcal{L} is a non-empty collection of subsets of X and I \subseteq X satisfies \left{x \in X : p_L(x) then p_I(x) = \sup \left{p_L(x) : L \in \mathcal{L}\right} for all x \in X.

The following examples show that the containment (0, R] K ; \subseteq ; {\textstyle\bigcap\limits_{e 0}} (0, R + e) K could be proper.

Example: If R = 0 and K = X then (0, R] K = (0, 0] X = \varnothing X = \varnothing but {\textstyle\bigcap\limits_{e 0}} (0, e) K = {\textstyle\bigcap\limits_{e 0}} X = X, which shows that its possible for (0, R] K to be a proper subset of {\textstyle\bigcap\limits_{e 0}} (0, R + e) K when R = 0. \blacksquare

The next example shows that the containment can be proper when R = 1; the example may be generalized to any real R 0. Assuming that [0, 1] K \subseteq K, the following example is representative of how it happens that x \in X satisfies p_K(x) = 1 but x \not\in (0, 1] K.

Example: Let x \in X be non-zero and let K = [0, 1) x so that [0, 1] K = K and x \not\in K. From x \not\in (0, 1) K = K it follows that p_K(x) \geq 1. That p_K(x) \leq 1 follows from observing that for every e 0, (0, 1 + e) K = [0, 1 + e)([0, 1) x) = [0, 1 + e) x, which contains x. Thus p_K(x) = 1 and x \in {\textstyle\bigcap\limits_{e 0}} (0, 1 + e) K. However, (0, 1] K = (0, 1]([0, 1) x) = [0, 1) x = K so that x \not\in (0, 1] K, as desired. \blacksquare

Positive homogeneity characterizes Minkowski functionals

The next theorem shows that Minkowski functionals are exactly those functions f : X \to [0, \infty] that have a certain purely algebraic property that is commonly encountered.

Let f : X \to [0, \infty] be any function. The following statements are equivalent:

1. Strict positive homogeneity: ; f(t x) = t f(x) for all x \in X and all positive real t 0.

• This statement is equivalent to: f(t x) \leq t f(x) for all x \in X and all positive real t 0.

1. f is a Minkowski functional: meaning that there exists a subset S \subseteq X such that f = p_S.
2. f = p_K where K := {x \in X : f(x) \leq 1}.
3. f = p_V , where V ,:= {x \in X : f(x)

Moreover, if f never takes on the value ,\infty, (so that the product 0 \cdot f(x) is always well-defined) then this list may be extended to include: |Positive/Nonnegative homogeneity: f(t x) = t f(x) for all x \in X and all nonnegative real t \geq 0.

If f(t x) \leq t f(x) holds for all x \in X and real t 0 then t f(x) = t f\left(\tfrac{1}{t}(t x)\right) \leq t \tfrac{1}{t} f(t x) = f(t x) \leq t f(x) so that t f(x) = f(t x).

Only (1) implies (3) will be proven because afterwards, the rest of the theorem follows immediately from the basic properties of Minkowski functionals described earlier; properties that will henceforth be used without comment. So assume that f : X \to [0, \infty] is a function such that f(t x) = t f(x) for all x \in X and all real t 0 and let K := {y \in X : f(y) \leq 1}. Note that if z \in X is such that f(z) \neq \infty then either z \in K or else f(z) 1 and \tfrac{1}{f(z)} z \in K so that either way, some positive real scalar multiple of z will belong to K. Similarly, if z \in X is such that f(z) \neq 0 then some positive real scalar multiple of z will not belong to K. It follows that f is identically ,\infty, (resp. identically 0) if and only if K = \varnothing (resp. K = X), which happens if and only if p_K is identically ,\infty, (resp. identically 0). So henceforth assume that K is neither empty nor all of X and that f is neither identically ,\infty, nor identically 0. --- END -----

For all real t 0, f(0) = f(t 0) = t f(0) so by taking t = 2 for instance, it follows that either f(0) = 0 or f(0) = \infty. Let x \in X. It remains to show that f(x) = p_K(x).

It will now be shown that if f(x) = 0 or f(x) = \infty then f(x) = p_K(x), so that in particular, it will follow that f(0) = p_K(0). So suppose that f(x) = 0 or f(x) = \infty; in either case f(t x) = t f(x) = f(x) for all real t 0. Now if f(x) = 0 then this implies that that t x \in K for all real t 0 (since f(t x) = 0 \leq 1), which implies that p_K(x) = 0, as desired. Similarly, if f(x) = \infty then t x \not\in K for all real t 0, which implies that p_K(x) = \infty, as desired. Thus, it will henceforth be assumed that R := f(x) a positive real number and that x \neq 0 (importantly, however, the possibility that p_K(x) is 0 or ,\infty, has not yet been ruled out).

Recall that just like f, the function p_K satisfies p_K(t x) = t p_K(x) for all real t 0. Since 0 p_K(x)= R = f(x) if and only if p_K\left(\tfrac{1}{R} x\right) = 1 = f\left(\tfrac{1}{R} x\right) so assume without loss of generality that R = 1 and it remains to show that p_K\left(\tfrac{1}{R} x\right) = 1. Since f(x) = 1, x \in K \subseteq (0, 1] K, which implies that p_K(x) \leq 1 (so in particular, p_K(x) \neq \infty is guaranteed). It remains to show that p_K(x) \geq 1, which recall happens if and only if x \not\in (0, 1) K. So assume for the sake of contradiction that x \in (0, 1) K and let 0 and k \in K be such that x = r k, where note that k \in K implies that f(k) \leq 1. Then 1 = f(x) = f(r k) = r f(k) \leq r \blacksquare

This theorem can be extended to characterize certain classes of [- \infty, \infty]-valued maps (for example, real-valued sublinear functions) in terms of Minkowski functionals. For instance, it can be used to describe how every real homogeneous function f : X \to \R (such as linear functionals) can be written in terms of a unique Minkowski functional having a certain property.

Characterizing Minkowski functionals on star sets

Let f : X \to [0, \infty] be any function and K \subseteq X be any subset. The following statements are equivalent:

1. f is (strictly) positive homogeneous, f(0) = 0, and {x \in X : f(x)
2. f is the Minkowski functional of K (that is, f = p_K), K contains the origin, and K is star-shaped at the origin.

• The set K is star-shaped at the origin if and only if t k \in K whenever k \in K and 0 \leq t \leq 1. A set that is star-shaped at the origin is sometimes called a star set.

Characterizing Minkowski functionals that are seminorms

In this next theorem, which follows immediately from the statements above, K is not assumed to be absorbing in X and instead, it is deduced that (0, 1) K is absorbing when p_K is a seminorm. It is also not assumed that K is balanced (which is a property that K is often required to have); in its place is the weaker condition that (0, 1) s K \subseteq (0, 1) K for all scalars s satisfying |s| = 1. The common requirement that K be convex is also weakened to only requiring that (0, 1) K be convex.

Let K be a subset of a real or complex vector space X. Then p_K is a seminorm on X if and only if all of the following conditions hold:

1. (0, \infty) K = X (or equivalently, p_K is real-valued).
2. (0, 1) K is convex (or equivalently, p_K is subadditive).

• It suffices (but is not necessary) for K to be convex.

1. (0, 1) u K \subseteq (0, 1) K for all unit scalars u.

• This condition is satisfied if K is balanced or more generally if u K \subseteq K for all unit scalars u.

in which case 0 \in K and both (0, 1) K = {x \in X : p(x) and \bigcap_{e 0} (0, 1 + e) K = \left{x \in X : p_K(x) \leq 1\right} will be convex, balanced, and absorbing subsets of X.

Conversely, if f is a seminorm on X then the set V := {x \in X : f(x) satisfies all three of the above conditions (and thus also the conclusions) and also f = p_V; moreover, V is necessarily convex, balanced, absorbing, and satisfies (0, 1) V = V = [0, 1] V.

If K is a convex, balanced, and absorbing subset of a real or complex vector space X, then p_K is a seminorm on X.

Positive sublinear functions and Minkowski functionals

It may be shown that a real-valued subadditive function f : X \to \R on an arbitrary topological vector space X is continuous at the origin if and only if it is uniformly continuous, where if in addition f is nonnegative, then f is continuous if and only if V := {x \in X : f(x) is an open neighborhood in X. If f : X \to \R is subadditive and satisfies f(0) = 0, then f is continuous if and only if its absolute value |f| : X \to 0, \infty) is continuous.

A nonnegative [sublinear function is a nonnegative homogeneous function f : X \to 0, \infty) that satisfies the triangle inequality. It follows immediately from the results below that for such a function f, if V := {x \in X : f(x) then f = p_V. Given K \subseteq X, the Minkowski functional p_K is a sublinear function if and only if it is real-valued and subadditive, which is happens if and only if (0, \infty) K = X and (0, 1) K is convex.

Correspondence between open convex sets and positive continuous sublinear functions

Suppose that X is a [topological vector space (not necessarily locally convex or Hausdorff) over the real or complex numbers. Then the non-empty open convex subsets of X are exactly those sets that are of the form z + {x \in X : p(x) for some z \in X and some positive continuous sublinear function p on X.

Let V \neq \varnothing be an open convex subset of X. If 0 \in V then let z := 0 and otherwise let z \in V be arbitrary. Let p = p_K : X \to [0, \infty) be the Minkowski functional of K := V - z where this convex open neighborhood of the origin satisfies (0, 1) K = K. Then p is a continuous sublinear function on X since V - z is convex, absorbing, and open (however, p is not necessarily a seminorm since it is not necessarily absolutely homogeneous). From the properties of Minkowski functionals, we have p_K^{-1}([0, 1)) = (0, 1) K, from which it follows that V - z = {x \in X : p(x) and so V = z + {x \in X : p(x) Since z + {x \in X : p(x) this completes the proof. \blacksquare

Notes

References